Calculus

DEMTH137

Edited by
Dr. Kulwinder Singh
 Calculus
Edited By:
Dr. Kulwinder Singh
 CONTENT

Unit 1: Sequence of Real Numbers 1

Dr. Pratiksha, Lovely Professional University

Unit 2: Definite integral as a limit of sum 10

Dr. Pratiksha, Lovely Professional University

Unit 3: Integration by using reduction formula 20

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Unit 4: Limit of a Real Valued Function 30

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Unit 5: Continuity of a real valued function 38
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Unit 6: Differentiability of a Real Valued Function 51

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Unit 7: Differentiability of a Real Valued Function 65
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Unit 8: Mean Value Theorems 73
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Unit 9: Higher Order Derivatives 81

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Unit 10: Maclaurin’s and Taylor’s Theorems 90
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Unit 11: Maxima and Minima of a Function 103
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Unit 12: Curvature and Asymptotes 113
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Unit 13: Concavity and Multiple Points 128
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Unit 14: Tracing of Curves 144

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 Notes
Dr. Pratiksha, Lovely Professional University Unit 01: Sequence of Real Numbers

Unit 1: Sequence of Real Numbers

CONTENTS
Objectives
Introduction
1.1 Real numbers
1.2 Number System
1.3 Sequence of real number
1.4 Bounded and unbounded sequence
1.5 Limit of a sequence
1.6 Convergence of sequences
1.7 Infinite series
1.8 Monotonic sequences
Summary
Key words
Self Assessment
Review Questions
Further/Suggested Readings

Objectives
Students will be able to

• cite some properties of real numbers.

• discuss the sequence of real numbers.
• explain the concept of convergence of a sequence.
• validate the bounded and monotonic sequences

Introduction
You have the idea about the number system since you began your journey of learning mathematics.
In this unit we will learn about some more properties of real numbers, the concept of sequence of real
numbers and their properties of being bounded and of convergence.

1.1 Real numbers

All around us, the one thing which is constant is the change. The study of change is called
CALCULUS. It has applications in almost all the fields of science and social science. For
understanding the change, one needs to have the concept of measurement. We need to measure the
time, distance, heat, force, intelligence etc. in order to adjudge the change happening in them.
One way to judge the change in any situation is by finding the average rate of change over a range.
If the change is to be judged at a point, then instantaneous rate of change is required. To understand
the instantaneous rate of change, one needs to understand the concept of limit first. Moreover, to
understand anything and everything one needs to understand the number system first.
The advent of Calculus is accredited to two mathematicians. One is British mathematician Sir Issac
Newton and other is German Mathematician Gottfried Leibnitz.

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Notes

1.2 Number System

We know that since ancient times, the pebbles were used by the shepherds etc. to count their sheep
and this amounts to the counting numbers, which are now known as natural numbers.
With the invention of zero, natural numbers {1, 2, 3,…} along with {0} are known as whole numbers.
Then the idea of negative numbers got evolved and the set {…, -2, -1, 0, 1, 2,…} was named as the set
of integers.
Then the idea of rational numbers came up where a rational number takes the form p/q where q is
non-zero number and p and q are integers. The numbers which are not rational were called irrational
numbers.
The rational and irrational numbers taken together are called real numbers. Therefore all the
terminating, nonterminating-repeating, nonterminating-nonrepeating decimals are the real
numbers.
All the real numbers can be plotted on the real line.

Closed and Open interval

Let 𝑎 and 𝑏 be two given numbers such that 𝑎 < 𝑏. Then, the set {𝑥: 𝑎 ≤ 𝑥 ≤ 𝑏} is called a closed
interval and is denoted by [𝑎, 𝑏]; the set {𝑥: 𝑎 < 𝑥 < 𝑏} is called a closed interval and is denoted by
(𝑎, 𝑏) and the set {𝑥: 𝑎 ≤ 𝑥 < 𝑏} or {𝑥: 𝑎 < 𝑥 ≤ 𝑏} is called a closed interval and is denoted by [𝑎, 𝑏) or
(𝑎, 𝑏] respectively.

Absolute function
Let 𝑥 be a real number then there exist three possibilities, then it can be more than, less than or equal
to zero. The modulus value or absolute value of 𝑥 is defined as

𝑥 𝑖𝑓 𝑥 ≥ 0
|𝑥| = { ∀𝑥 ∈ℝ
−𝑥 𝑖𝑓 𝑥 < 0
Moreover, we can write |𝑥| = max {𝑥, −𝑥}.

Also |𝑥 − 𝑎| > 𝑙 ⇒ x − a > l or x − a < −l and

|𝑥 − 𝑎| < 𝑙 ⇒ −l < x − a < l

1.3 Sequence of real number

2 LOVELY PROFESSIONAL UNIVERSITY

 Notes
Unit 01: Sequence of Real Numbers

Sequence is a kind of ordered list of numbers where a pattern can be seen. For example, 3, 6, 9, . . . is
a list in which the first term is 3, second is 6 and so on. We can easily tell that the fourth term will be
12 and so on. This example here is of an infinite sequence.

Technically, a function whose domain is a set of natural numbers and range is a subset of real
numbers, is called a real sequence (and just sequence in this course). Since the domain of all sequences
is the set of natural numbers, therefore a sequence is completely determined if
𝑓(𝑛) ∀ 𝑛 𝜖 𝑵 is known. The sequence 𝑓 is denoted as < 𝑓𝑛 > 𝑜𝑟 {𝑓𝑛 } mostly.

1
{ } , {(−1)𝑛 } etc.
𝑛

Range of sequence is the set of all distinct terms of a sequence. It can be infinite or finite. The first
sequence in the example is an infinite one and the second one is finite.

1.4 Bounded and unbounded sequence

A sequence is said to be bounded if and only if its range is bounded.

A sequence {𝑓𝑛 } is bounded above if there exists a real number K such that

𝑓𝑛 ≤ 𝐾 ∀ 𝑛𝜖𝑵

A sequence {𝑓𝑛 } is bounded below if there exists a real number k such that

𝑓𝑛 ≥ 𝑘 ∀ 𝑛𝜖𝑵

A sequence is said to be bounded if it is bounded above as well as below. Hence a sequence

{𝑓𝑛 } is said to be bounded if there exist two real numbers 𝑘 and 𝐾 such that

𝑘 ≤ 𝑓𝑛 ≤ 𝐾 ∀ 𝑛 𝜖 𝑵

We call the sequence to be unbounded if it is not bounded.

Prove that the sequence {n} is not convergent and is not bounded.

1 1
Let 𝑝 be a real number. Then, the neighbourhood (𝑝 − , 𝑝 + ) will contain at most one term of the
4 4
sequence {𝑛}. Therefore, in this neighbourhood, we cannot find infinitely many terms of the sequence.
Hence the sequence is not convergent.
On the other hand, we can see that all the elements of the sequence are more than 1 but we cannot
find any upper bound i.e. any number so that all the elements of the sequence are below that number.
Therefore, the sequence is just bounded below and not bounded above or simply the sequence is not
bounded.

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Notes

1.5 Limit of a sequence

This concept of approach is of utmost importance to understand the calculus. A number a is called
the limit of an infinite sequence {𝑓𝑛 } if for any positive number 𝜖 we can find a positive number 𝑵
depending on 𝜖 such that

|𝑓𝑛 − 𝑎| < 𝜖 ∀ integers 𝑛 > 𝑵

That is in any neighbourhood of 𝑎 , 𝑓𝑛 belongs to the same neighbourhood for infinite values of 𝑛.
Consider the sequence {𝑛}. If you consider any neighbourhood of any number 𝑎, it will not
contain an infinite number of elements of the sequence. On the other hand any neighbourhood of
1
any number a , for the sequence { } will contain infinitely many elements of the sequence.
𝑛

1.6 Convergence of sequences

A sequence is said to be convergent if its limit exists and is unique. In other words, a sequence {𝑓𝑛 }
has a limit 𝑎 if the successive terms get closer and closer to 𝑎.

Theorems on limits of sequence

𝑙𝑖𝑚 𝑙𝑖𝑚
If 𝑛→∞ 𝑓𝑛 = 𝐴 and 𝑛→∞ 𝑔𝑛 = 𝐵, then

𝑙𝑖𝑚
(i) 𝑛→∞ ( 𝑓𝑛 + 𝑔𝑛 ) = 𝐴 + 𝐵

𝑙𝑖𝑚
(ii) 𝑛→∞ ( 𝑓𝑛 − 𝑔𝑛 ) = 𝐴 − 𝐵

(iii) 𝑙𝑖𝑚 𝑓𝑛 . 𝑔𝑛 = 𝐴. 𝐵
𝑛→∞

4 LOVELY PROFESSIONAL UNIVERSITY

 Notes
Unit 01: Sequence of Real Numbers

𝑙𝑖𝑚 𝑓 𝐴
(iv) 𝑛→∞ ( 𝑛) = provided 𝐵 ≠ 0
𝑔𝑛 𝐵

If B=0, A≠0 then the limit does not exist and if B=0, A=0 then the limit may or may not exist.

𝑙𝑖𝑚 𝑝
(v) 𝑛→∞ 𝑓𝑛 = 𝐴𝑝 , 𝑝𝜖𝑹

(vi) 𝑙𝑖𝑚 𝑝 𝑓𝑛 = 𝑝 𝐴 , 𝑝𝜖𝑹

𝑛→∞

1.7 Infinite series

Let 𝑓1 , 𝑓2 , 𝑓3 , . . . be a sequence. Create a new sequence by taking the sum of the terms taken one,
two, three… at a time. Let 𝑆1 , 𝑆2 , 𝑆3 , . .. be such that

𝑆1 = 𝑓1
𝑆2 = 𝑓1 + 𝑓2
𝑆3 = 𝑓1 + 𝑓2 + 𝑓3
.
.
.
𝑆𝑛 = 𝑓1 + 𝑓2 + 𝑓3 + ⋯ + 𝑓𝑛
Here 𝑆𝑛 is called the 𝑛𝑡ℎ partial sum. The sequence 𝑆1 , 𝑆2 , 𝑆3 , . .. is symbolized by

𝑓1 + 𝑓2 + 𝑓3 + ⋯ = f
n =1
n

which is called an infinite series. If 𝑛→∞

𝑙𝑖𝑚 𝑆 = 𝑆 exists, the series is called convergent and 𝑆 is called
𝑛
its sum, otherwise the series is called divergent.
For instance, the geometric progression (G. P.) 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ . +𝑎𝑟 𝑛−1 + ⋯ is a series which
𝑎
converges to sum (1−𝑟) provided |𝑟| < 1 and diverges if |𝑟| ≥ 1.

1.8 Monotonic sequences

A sequence {𝑓𝑛 } is said to be monotonically increasing if 𝑓𝑛+1 ≥ 𝑓𝑛 ∀ 𝑛 ∈ 𝑵 and monotonically
decreasing if 𝑓𝑛+1 ≤ 𝑓𝑛 ∀ 𝑛 ∈ 𝑵. For example, the sequence {𝑛} monotonically increasing and the
1
sequence { } is monotonically decreasing. The sequence which is either monotonically increasing or
𝑛
decreasing is called the monotonic sequence.

Can you recognize what kind of sequence is being depicted by the following graphs?

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 Calculus
Notes

2𝑛−7
Prove that the sequence { } is (i) monotonically increasing (ii) is bounded and (iii)
3𝑛+2
2
tends to limit { }.
3
2𝑛−7
Let 𝑓𝑛 =
3𝑛+2

2𝑛−5
then 𝑓𝑛+1 =
3𝑛+5

2𝑛−5 2𝑛−7 25
And 𝑓𝑛+1 − 𝑓𝑛 = − = (3𝑛+5)(3𝑛+2) > 0 ∀ 𝑛 ∈ 𝑵
3𝑛+5 3𝑛+2

This shows that the given sequence is monotonically increasing. Now if we write few terms of the
sequence {𝑓𝑛 } , we can observe that all the terms are more than or equal to −1.

i.e. 𝑓𝑛 ≥ −1 ∀ 𝑛 ∈ 𝑵

𝑛+9
Moreover 1 − 𝑓𝑛 = > 0 ∀𝑛 ∈𝑵
3𝑛+2

This implies 𝑓𝑛 < 1 ∀ 𝑛 ∈ 𝑵

And −1 ≤ 𝑓𝑛 < 1 ∀ 𝑛 ∈ 𝑵 , therefore the sequence is bounded.

Now the limit can be found as follows,

7
2−
𝑙𝑖𝑚 𝑙𝑖𝑚 2𝑛−7 𝑙𝑖𝑚 𝑛 2
𝑛→∞ 𝑓𝑛 = 𝑛→∞ 3𝑛+2 = 𝑛→∞ 2 =3
3+
𝑛

Summary
In this chapter we have seen the concept of the sequence, its convergence, boundedness, and
monotonicity.

Key words
Real Numbers, Sequence, Series, Convergence, Bounded function, Monotonic function

Self Assessment

1. The interval [3,99) is a

(a) closed interval

(b) open interval
(c) semi closed or semi open interval

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 Notes
Unit 01: Sequence of Real Numbers

(d) semi open interval

2. |𝑥| can be written as

(a) 𝑥
(b) −𝑥
(c) min{−𝑥, 𝑥}
(d) max {−𝑥, 𝑥}

3. The solution of |𝑥 − 5| < 3 can be written as

(a) [2,8]
(b) (2,8)
(c) (2,8]
(d) (-2,8)

4. The solution of |𝑥 − 5| ≥ 3 can be written as

(a) (−∞, 2]𝑈[8, ∞)

(b) [2,8]
(c) (−∞, 2)𝑈(8, ∞)
(d) None of these

5. Which of the following is an infinite sequence?

(a) The prime numbers between 2 to 2000

(b) The set of even numbers
(c) The set of odd numbers between 3 to 30
(d) First ten multiples of seven

6. Function f is bounded if its range f(A) is a __________ subset.

7. The supremum of a set is its least upper bound.

True

False

8. The infimum of a set is its greatest upper bound.

True

False

9. The sequence {𝑛} is

(a) not convergent

(b) not bounded

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 Calculus
Notes

(c) convergent and bounded

(d) neither convergent nor bounded

10. A sequence is said to be convergent

(a) if its limit exists

(b) if it is unique
(c) if its limit exists and is unique
(d) None of these

𝑙𝑖𝑚 𝑙𝑖𝑚
11. If 𝑛→∞ 𝑓𝑛 = 𝐴 and 𝑛→∞ 𝑔𝑛 = 𝐵, then

(a) 𝑙𝑖𝑚 ( 𝑓𝑛 + 𝑔𝑛 ) = 𝐴 + 𝐵
𝑛→∞

(b) 𝑙𝑖𝑚 ( 𝑓𝑛 − 𝑔𝑛 ) = −𝐴 − 𝐵
𝑛→∞

(c) 𝑙𝑖𝑚 𝑓𝑛 . 𝑔𝑛 = 𝐴 + 𝐵
𝑛→∞

(d) 𝑙𝑖𝑚 𝑓𝑛 . 𝑔𝑛 = 𝐴/𝐵

𝑛→∞

2𝑛−7
12. The sequence { } converges to
3𝑛+2

(a) 2/3
(b) -1
(c) 1
(d) none of these

13. A sequence {𝑓𝑛 } is said to be monotonically increasing if

(a) 𝑓𝑛+1 ≤ 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(b) 𝑓𝑛+1 < 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(c) 𝑓𝑛+1 > 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(d) 𝑓𝑛+1 ≥ 𝑓𝑛 ∀ 𝑛 ∈ 𝑁

14. A sequence {𝑓𝑛 } is said to be strictly monotonically increasing if

(a) 𝑓𝑛+1 ≤ 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(b) 𝑓𝑛+1 < 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(c) 𝑓𝑛+1 > 𝑓𝑛 ∀ 𝑛 ∈ 𝑁
(d) 𝑓𝑛+1 ≥ 𝑓𝑛 ∀ 𝑛 ∈ 𝑁

15. The sequence {𝑛} is

(a) convergent
(b) monotonic
(c) convergent and monotonic
(d) neither convergent nor monotonic

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 Notes
Unit 01: Sequence of Real Numbers

Answers:

1 c 6 bounded 11 a

2 d 7 True 12 a

3 b 8 True 13 d

4 a 9 d 14 c

5 b 10 c 15 B

Review Questions
1. What is the solution of |𝑥 − 1| < 5 ?
2. Check whether the sequence {(−𝑛)𝑛 } is bounded or not?
3. What is the solution of |𝑥 − 2| < 6 ?
4. What is the solution of |𝑥 − 1| > 5 ?
5. Prove that |𝑎𝑏| = |𝑎||𝑏|.
6. State true or false: {𝑥: |𝑥 − 3| < 4} = {𝑥: −1 < 𝑥 < 7}
𝑙𝑖𝑚 𝑛
7. Evaluate 𝑛→∞ log 5
log9 (𝑛)

8. Check if the sequence {(−2)𝑛 } is bounded or not?

2
9. Check if the sequence { } is bounded or not?
𝑛

10. Check if the sequence {𝑛2 } is bounded or not?

11. Write a short note on boundedness of a sequence.
2
12. Check the monotonicity of the sequence { }.
𝑛
2𝑛
13. Check the monotonicity of the sequence { }.
𝑛+1

14. Write a short note on the monotonicity of a sequence.

15. Write a short note on the convergence of a sequence.

Further/Suggested Readings
1. George B. Thomas Jr., Joel Hass, Christopher Heil & Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
2. Howard Anton, I. Bivens & Stephan Davis (2016). Calculus (10th edition). Wiley India.
3. http://mathonline.wikidot.com/calculus

LOVELY PROFESSIONAL UNIVERSITY 9

Dr. Pratiksha, Lovely Professional University Unit 02: Definite integral as a limit of sum

Notes

Unit 02: Definite integral as a limit of sum

CONTENTS
Objectives
Introduction
2.1 Integral as a limit of sum
2.2 Integration of irrational algebraic functions
2.3 Integration of transcendental functions
Summary
Key words
Review Questions
Further/Suggested Readings

Objectives
Students will -

• learn the concept of integral as a limit of sum.

• learn about the hyperbolic functions
• be able to integrate irrational algebraic functions.
• be able to integrate the transcendental functions

Introduction
You have learnt the rules of differentiation and integration at the senior secondary level. The definite
integral of a function gives you the area under the curve of that function between the specified limits.
In this unit we will look into the detail as to what is the integral as a limit of sum. We will evaluate
some definite integrals with this ab initio method. The functions can be classified as algebraic and
transcendental functions. The polynomial functions, rational functions are the algebraic functions
and exponential, logarithmic, trigonometric, inverse trigonometric functions, hyperbolic functions,
inverse hyperbolic functions are the examples of the transcendental functions. We will solve some
problems on integration of irrational as well as transcendental functions in this unit.

2.1 Integral as a limit of sum

Consider a continuous function defined on a closed interval [a, b], where all the values of the function
are non-negative. The area bound between the curve, the point 𝑥 = 𝑎 and 𝑥 = 𝑏 and the x-axis is the
𝑏
definite integral ∫𝑎 𝑓(𝑥)𝑑𝑥 of any such continuous function 𝑓.

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Notes First idea is to divide the interval [𝑎, 𝑏] onto 𝑛 equal sub-intervals as:
𝑏−𝑎
𝑎 = 𝑥0 , 𝑥1 , 𝑥2 , … , 𝑥𝑛−2 , 𝑥𝑛−1 , 𝑥𝑛 = 𝑏 . Now we can see that the length of each sub interval must be
𝑛

and let it be denoted by ℎ. Therefore we can write the points of the partition as follows:
𝑥0 = 𝑎,
𝑥1 = 𝑎 + ℎ
𝑥2 = 𝑎 + 2ℎ
.
.
.
𝑥𝑛 = 𝑎 + 𝑛ℎ = 𝑏
Clearly as → ∞ , ℎ → 0. In the above figure The region PRSQP is the sum of all the 𝑛 sub-regions,
where each sub-region is defined on sub-interval [𝑥𝑟−1 , 𝑥𝑟 ], 𝑟 = 1, 2, 3, … 𝑛. Observe region ABDM.
Area of the rectangle (ABLC) < Area of the region (ABDCA) < Area of the rectangle (ABDM).
As ℎ → 0 all these areas become almost equal to each other. Hence, we have
𝑛−1

𝑠𝑛 = ℎ[𝑓(𝑥0 ) + 𝑓(𝑥1 ) + ⋯ + 𝑓(𝑥𝑛−1 )] = ℎ ∑ 𝑓(𝑥𝑟 )

𝑟=0

and
𝑛

𝑆𝑛 = ℎ[𝑓(𝑥1 ) + 𝑓(𝑥2 ) + ⋯ + 𝑓(𝑥𝑛 )] = ℎ ∑ 𝑓(𝑥𝑟 )

𝑟=1

Here 𝑠𝑛 and 𝑆𝑛 denote the sum of areas of all lower rectangles and upper rectangles raised over
subintervals [𝑥𝑟−1 , 𝑥𝑟 ], 𝑟 = 1, 2, 3, … 𝑛 respectively.
Therefore we can write
𝑠𝑛 < area of the region < 𝑆𝑛
As 𝑛 → ∞, these strips become narrower. Further it is assumed that the limiting value of 𝑠𝑛 and 𝑆𝑛
are the same in both cases and the common limiting value is the required area under the curve.
Symbolically we can write
𝑏
𝑙𝑖𝑚 𝑠𝑛 = 𝑙𝑖𝑚 𝑆𝑛 =area (PRSQP) =∫𝑎 𝑓(𝑥)𝑑𝑥
𝑛→∞ 𝑛→∞

This area is also the limiting value of any area which is between that of the rectangles below the curve
and that of the rectangles above the curve. For convenience, we shall take the rectangles having
height equal to that of the curve at the left-hand-edge of each sub- interval. Hence, we can write
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 𝑙𝑖𝑚 ℎ[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)]
𝑎 𝑛→∞

𝑏 (𝑏 − 𝑎) 𝑙𝑖𝑚 [𝑓(𝑎) + 𝑓(𝑎 + ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)]

𝑛→∞
∫ 𝑓(𝑥)𝑑𝑥 =
𝑎 𝑛
𝑏−𝑎
where ℎ = → 0 𝑎𝑠 𝑛 → ∞. This equation is the definition of definite integral as the limit of a sum.
𝑛

5
Evaluate ∫2 𝑥 2 𝑑𝑥 as a limit of sum.

Here 𝑓(𝑥) = 𝑥 2
𝑏−𝑎 3
𝑎 = 2, 𝑏 = 5, ℎ = =
𝑛 𝑛

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 Unit 02: Definite integral as a limit of sum

𝑏
Notes
∫ 𝑓(𝑥)𝑑𝑥 = 𝑙𝑖𝑚 ℎ[𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ) + ⋯ + 𝑓(𝑎 + 𝑛ℎ)]
𝑎 𝑛→∞

5
3
∫ 𝑥 2 𝑑𝑥 = 𝑙𝑖𝑚 [(2 + ℎ)2 + (2 + 2ℎ)2 + ⋯ + (2 + 𝑛ℎ)2 ]
2 𝑛→∞ 𝑛

3
= 𝑙𝑖𝑚 [𝑛 ⋅ 22 + ℎ2 (12 + 22 + ⋯ + 𝑛2 ) + 4ℎ(1 + 2 + ⋯ 𝑛)]
𝑛→∞ 𝑛

3 𝑛(𝑛 + 1)(2𝑛 + 1) 𝑛(𝑛 + 1)

= 𝑙𝑖𝑚 [𝑛 ⋅ 22 + ℎ2 ( ) + 4ℎ ( )]
𝑛→∞ 𝑛 6 2

3 3 2 𝑛(𝑛 + 1)(2𝑛 + 1) 3 𝑛(𝑛 + 1)

= 𝑙𝑖𝑚 [𝑛 ⋅ 22 + ( ) ( ) + 4( )( )]
𝑛→∞ 𝑛 𝑛 6 𝑛 2

3 1 1 2 1
3 2
3 2 𝑛 (1 + 𝑛) (2 + 𝑛) 3 𝑛 (1 + 𝑛)
= 𝑙𝑖𝑚 [𝑛 ⋅ 2 + ( ) ( ) + 4( )( )]
𝑛→∞ 𝑛 𝑛 6 𝑛 2

1 1 1
(1 + ) (2 + ) (1 + )
= 𝑙𝑖𝑚 [3. 22 + 3(3)2 ( 𝑛 𝑛 ) + 3(4)(3) ( 𝑛 )]
𝑛→∞ 6 2

= 12 + 9 + 18 = 39
Thus the limit of the sum or in other words the area under the curve 𝑥 2 between lines 𝑥 = 2 and 𝑥 =
5 and above the 𝑥-axis is 39 square units.

𝑏
Evaluate ∫𝑎 cos 𝑥 𝑑𝑥 as the limit of a sum.

Here 𝑓(𝑥) = cos 𝑥

By definition of integral as a limit of sum we can write
𝑏
∫𝑎 cos 𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ℎ[cos(𝑎 + ℎ) + cos(𝑎 + 2ℎ) + cos(𝑎 + 3ℎ) + ⋯ + cos(𝑎 + 𝑛ℎ)]
𝑛→∞

Let 𝑆 = cos(𝑎 + ℎ) + cos(𝑎 + 2ℎ) + cos(𝑎 + 3ℎ) + ⋯ + cos(𝑎 + 𝑛ℎ)

ℎ
To calculate this sum, we must multiply both sides 2 sin ( ). We get,
2

ℎ ℎ
2 sin ( ) 𝑆 = 2 sin ( ) ( cos(𝑎 + ℎ) + cos(𝑎 + 2ℎ) + cos(𝑎 + 3ℎ) + ⋯ + cos(𝑎 + 𝑛ℎ)
2 2
3 1 5 3 2𝑛+1
= sin (𝑎 + ℎ) − sin (𝑎 + ℎ) + sin (𝑎 + ℎ) − sin (𝑎 + ℎ) + … + sin (𝑎 + ℎ) −
2 2 2 2 2
2𝑛−1
sin (𝑎 + ℎ)
2
2𝑛+1 1
= sin (𝑎 + ℎ) − sin (𝑎 + ℎ)
2 2
1 1
= sin (𝑏 + ℎ) − sin (𝑎 + ℎ) as 𝑛ℎ = 𝑏 − 𝑎
2 2
1 1
𝑏 sin(𝑏+ ℎ)−sin(𝑎+ ℎ)
Thus ∫𝑎 cos 𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ℎ 2
ℎ
2
𝑛→∞ 2 sin( )
2

= sin 𝑏 − sin 𝑎
In the senior secondary level, you have learnt how to integrate the algebraic rational functions and
the trigonometric functions. In the next section we will see some cases of the integration of irrational
algebraic function.

𝑏 1
Evaluate ∫𝑎 𝑑𝑥 as limit of sum.
√𝑥

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 Calculus

Notes
2.2 Integration of irrational algebraic functions
Certain types of integrals containing irrational expressions can be reduced to integrals of rational
functions by making appropriate substitutions. These substitutions are done with an intention to
convert the irrational function into a rational one.

𝑚
To integrate a function that contains only one irrational expression of the form 𝑥 𝑛 we
1
make the substitution for 𝑥 𝑛 .

1 1
𝑎𝑥+𝑏 𝑛 𝑎𝑥+𝑏 𝑛
An expression of the form ( ) can be integrated by using the substitution for ( ) , where
𝑐𝑥+𝑑 𝑐𝑥+𝑑
𝑎, 𝑏, 𝑐, 𝑑 are real numbers. These substitutions reduce the integrals rational functions in the
transformed variable.

The integrals containing radicals of the form √𝑎2 − 𝑥 2 , √𝑎2 + 𝑥 2 and √𝑥 2 − 𝑎2 can be evaluated with
the help of trigonometric and hyperbolic substitutions.

𝑥
Consider the integral of ∫ 1 𝑑𝑥
(𝑎+𝑏𝑥)3

The denominator is having an irrational function and if we substitute 𝑎 + 𝑏𝑥 = 𝑦 3 then the given
integrand can be written as a rational function in 𝑦. Let us see the steps.
If we substitute 𝑎 + 𝑏𝑥 = 𝑦 3
Then 𝑏 𝑑𝑥 = 3𝑦 2 𝑑𝑦
𝑥 𝑦 3 −𝑎 3𝑦 2
And ∫ 1 𝑑𝑥 = ∫ ( )( ) 𝑑𝑦
(𝑎+𝑏𝑥)3 𝑏 𝑦𝑏

1
= ∫ (𝑦 4 − 𝑎𝑦)𝑑𝑦
𝑏2
3 𝑦5 𝑎𝑦 2
= [ − ]+C
𝑏2 5 2
3
= 𝑦 2 (2𝑦 3 − 5𝑎) + 𝐶
10𝑏2

where 𝐶 is the constant of integration.

√𝑥+9
Evaluate ∫ ( ) 𝑑𝑥
𝑥

Put √𝑥 + 9 = 𝑢
𝑑𝑥 = 2 𝑢 𝑑𝑢
Then
√𝑥+9 𝑢 𝑢2
∫( ) 𝑑𝑥 = ∫ ( ) 2𝑢𝑑𝑢 = 2 ∫ 𝑑𝑢
𝑥 𝑢2 −9 𝑢2 −9
𝑢2 −9+9
= 2∫ 𝑑𝑢-
𝑢2 −9

9
= 2 ∫ 𝑑𝑢 + 2∫ 𝑑𝑢
𝑢2 − 9
1 𝑢−3
= 2 𝑢 + 2. 9. ln | |+𝐶
6 𝑢+3
= 2 √𝑥 + 9 + 3 ln |(√𝑥 + 9 − 3)/(√𝑥 + 9 + 3) + 𝐶
where 𝐶 is the constant of integration.

1
Evaluate ∫ (5𝑥 − 1)3 𝑑𝑥

1
To solve this integral, substitute (5𝑥 − 1)3 = 𝑢

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 Unit 02: Definite integral as a limit of sum

Notes
(5𝑥 − 1) = 𝑢3
5 𝑑𝑥 = 3 𝑢2 𝑑𝑢
Then
1 3𝑢2
∫ (5𝑥 − 1)3 𝑑𝑥 = ∫ 𝑢 𝑑𝑢
5
3 4
= 𝑢 +𝐶
20
3 4
= (5𝑥 − 1)3 + 𝐶
20
2.3 Integration of transcendental functions
A number that is not the root of any integer polynomial is termed as a transcendental number. And
on the same lines a function that cannot be written using roots and the arithmetic found in
polynomials is known as a transcendental function.
e.g. Exponential function
Logarithmic function
Trigonometric function
Inverse trigonometric function
Hyperbolic function
Inverse hyperbolic functions etc.

You are familiar with all the above mentioned functions except the hyperbolic
function. So here is a short introduction to this family of transcendental functions.

Hyperbolic functions are the functions defined in terms of the exponential functions as follows.

Just like with trigonometric functions, there are identities related to the hyperbolic functions.

Rules of differentiation and integration are as follows. Here the prime symbol is for denoting the first
derivative.

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 Calculus

Notes

The graphs of the hyperbolic functions are shown below and it can be seen that with appropriate
range restrictions, they all have inverses (same as the case with the inverse trigonometric functions).

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 Unit 02: Definite integral as a limit of sum

Notes

The range in case of sec ℎ−1 𝑥 is [0, ∞) and for coth−1 𝑥 is (−∞, 0)𝑈(0, ∞). The derivative of the
inverse hyperbolic functions are as follows:

The integral as an anti-derivative can be written easily from the above rules of differentiation.

16 LOVELY PROFESSIONAL UNIVERSITY

 Calculus

Notes Thus you got a brief idea about the calculus of hyperbolic functions and inverse hyperbolic functions.
For more information you can check the following link.

https://math.libretexts.org/Courses/Monroe_Community_College/MTH_21
1_Calculus_II/Chapter_6%3A_Applications_of_Integration/6.9%3A_Calculus
_of_the_Hyperbolic_Functions

Now with the knowledge of the transcendental functions, we can look into the examples related to
the integration of the transcendental functions.

𝑥𝑒 𝑥
Evaluate ∫ (𝑥+1)2
𝑑𝑥

𝑥𝑒 𝑥 (𝑥 + 1 − 1)𝑒 𝑥
∫ 𝑑𝑥 = ∫ 𝑑𝑥
(𝑥 + 1)2 (𝑥 + 1)2
1 1
=∫ ( − (𝑥+1)2) 𝑒 𝑥 𝑑𝑥
𝑥+1

𝑒𝑥
= +𝐶 (Using integration by parts on the first term)
𝑥+1

Evaluate ∫ sin (log 𝑥) 𝑑𝑥

Put log 𝑥 = 𝑡
1
i.e. 𝑑𝑥 = 𝑑𝑡
𝑥

𝑑𝑥 = 𝑒 𝑡 𝑑𝑡
Therefore ∫ sin (log 𝑥) 𝑑𝑥 = ∫ 𝑒 𝑡 sin𝑡 𝑑𝑡
= −𝑒 −𝑡 cos 𝑡 + 𝑒 𝑡 sin 𝑡 − ∫ 𝑒 𝑡 sin𝑡 𝑑𝑡
Clubbing the integral term, we get
2∫ 𝑒 𝑡 sin 𝑡 𝑑𝑡 = 𝑒 𝑡 (sin 𝑡 − cos 𝑡) + 𝐶
𝑒𝑡 𝐶
∫ 𝑒 𝑡 sin 𝑡 𝑑𝑡 = (sin 𝑡 − cos 𝑡) +
2 2
1
∫ sin (log 𝑥) 𝑑𝑥 = 𝑥(sin (log 𝑥) − cos (𝑙𝑜𝑔𝑥)) + 𝑪
2

Calculate 𝐼 = ∫ 𝑒 𝑎𝑥 cos (𝑏𝑥 + 𝑐)𝑑𝑥

𝑒 𝑎𝑥 𝑏 𝑎𝑥
𝐼= cos (𝑏𝑥 + 𝑐) + ∫ ( ) sin (𝑏𝑥 + 𝑐) 𝑒 𝑑𝑥
𝑎 𝑎

𝑒 𝑎𝑥 𝑏 𝑎𝑥 𝑏2 𝑎𝑥
𝐼= cos (𝑏𝑥 + 𝑐) + (𝑎2) sin (𝑏𝑥 + 𝑐) 𝑒 − 𝑎2 ∫ 𝑐𝑜𝑠(𝑏𝑥 + 𝑐) 𝑒 𝑑𝑥 (integrating by parts)
𝑎

𝑏2 𝑒 𝑎𝑥
(1 + ) 𝐼 = (𝑎 cos(𝑏𝑥 + 𝑐) + 𝑏 sin (𝑏𝑥 + 𝑐))
𝑎2 𝑎2
𝑒 𝑎𝑥
𝐼= ( 𝑎 cos(𝑏𝑥 + 𝑐) + 𝑏 sin(𝑏𝑥 + 𝑐))
𝑎2 + 𝑏2

1−𝑥
Integrate tan−1 √ with respect to 𝑥.
1+𝑥

1−𝑥
Let 𝐼 = tan−1 √
1+𝑥

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 Unit 02: Definite integral as a limit of sum

Notes
Put 𝑥 = cos 𝜃
𝑑𝑥 = − sin 𝜃 𝑑𝜃

Therefore 𝐼 = tan−1 √[(1 − cos 𝜃)/(1 + cos 𝜃)]

𝜃
= tan−1 tan ( )
2
𝜃
=
2
1−𝑥 𝜃
Now ∫ tan−1 √ 𝑑𝑥 = ∫ (− sin 𝜃)𝑑𝜃
1+𝑥 2

1
= − (−𝜃 cos 𝜃 + 𝑠𝑖𝑛𝜃)
2
1
= − (−𝑥 cos −1 𝑥 + √1 − 𝑥 2 ) + 𝐶
2

𝑠𝑖𝑛ℎ𝑥
Evaluate ∫ 𝑑𝑥
1+𝑐𝑜𝑠ℎ𝑥

For this kind of integral we can go for a suitable substitution.

Let 1 + 𝑐𝑜𝑠ℎ𝑥 = 𝑢
It gives 𝑠𝑖𝑛ℎ𝑥 𝑑𝑥 = 𝑑𝑢
𝑠𝑖𝑛ℎ𝑥 𝑑𝑢
∫ 𝑑𝑥 = ∫ = ln|𝑢| + 𝐶 = ln |1 + cosh 𝑥| + 𝐶
1 + 𝑐𝑜𝑠ℎ𝑥 𝑢

Evaluate 𝐼 = ∫ cos 𝑥/√(1 + sin2 𝑥) 𝑑𝑥

Put sin 𝑥 = 𝑢
cos 𝑥 𝑑𝑥 = 𝑑𝑢
1
𝐼=∫ 𝑑𝑢 (Refer to the rules of differentiation (integration) of the inverse hyperbolic functions)
√1+𝑢2

= sinh−1 𝑢 + 𝐶
= sinh−1 sin 𝑥 + 𝐶

1
Evaluate the integral ∫ 𝑑𝑥
√4𝑥 2 −1

Let us substitute 2𝑥 = 𝑢
2𝑑𝑥 = 𝑑𝑢 and this results in
1 1
∫ 𝑑𝑥 = ∫ 𝑑𝑢
√4𝑥 2 −1 2√𝑢2 − 1
1
= cosh−1 𝑢 + 𝐶
2
1
= cosh−1 (2𝑥) + 𝐶
2
Summary
In this chapter we have seen how a definite integral can be calculated as a limit of sum. This is the ab
initio way to calculate the area under a curve 𝑦 = 𝑓(𝑥) bounded by the two vertical lines 𝑥 = 𝑎, 𝑥 =
𝑏 and the x-axis.

Key words
Real Numbers, Sequence, Series, Convergence, Bounded function, Monotonic function

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 Calculus

Notes
Review Questions
Evaluate the following:

1. ∫ sinh3 𝑥 cosh 𝑥 𝑑𝑥
2. ∫ sech2 3𝑥 𝑑𝑥
𝑑
3. (cosh−1 3𝑥)
𝑑𝑥
1
4. ∫ √1−𝑒 2𝑥 𝑑𝑥
5. ∫ √𝑒 𝑥 + 1𝑑𝑥
𝑥 2 −2
6. ∫ 𝑑𝑥
𝑥+1
1−𝑥 𝑑𝑥
7. ∫
1+𝑥 𝑥
8. ∫ √𝑠𝑒𝑐𝑥 − 1 𝑑𝑥
𝑑𝑥
9. ∫
5+4𝑐𝑜𝑠𝑥
10. ∫ 𝑐𝑜𝑠𝑒𝑐 5 𝑥 𝑑𝑥
11. ∫ sinh3 𝑥 𝑑𝑥
12. ∫ 𝑥 sinh 𝑥 𝑑𝑥
3
13. Evaluate the definite integral ∫2 𝑥 3 𝑑𝑥 as limit of sum.
𝑏
14. Evaluate the definite integral ∫𝑎 sinh 𝑥 𝑑𝑥 as limit of sum.
𝜋/2
15. Evaluate the definite integral ∫0 cos 𝑥 𝑑𝑥 as limit of sum.

Further/Suggested Readings
George B. Thomas Jr., Joel Hass, Christopher Heil & Maurice D. Weir (2018). Thomas’ Calculus
(14th edition). Pearson Education.
Howard Anton, I. Bivens & Stephan Davis (2016). Calculus (10th edition). Wiley India.
Shanti Narayan & Dr. P. K. Mittal (2018). Integral Calculus (11th edition). S. Chand

LOVELY PROFESSIONAL UNIVERSITY 19

Dr. Pratiksha, Lovely Professional University Unit 03: Integration by using reduction formula
Notes

Unit 03: Integration by using reduction formula

CONTENTS
Objectives
Introduction
3.1 Reduction formula
3.2 Properties of definite integral
Summary
Key words
Self Assessment
Review Questions
Further/Suggested Readings

Objectives
Students will

• learn about an alternate method to solve the integrals.

• be able to solve the problems employing reduction formula.
• prove various properties of definite integrals.
• be able to solve definite integrals by using its properties.

Introduction
By now, you are well aware of the methods of substitution, method of integration by parts and the
method in which we decompose the given integrand in the sum of integrands with known integrals.
There is one more technique and that is called integration by successive reduction or integration
using reduction formula.

3.1 Reduction formula

Any formula which expresses an integral in terms of another integral of the same type but of lesser
degree or order is called a reduction formula. The successive application of the reduction formula
enables us to express the integral of the general member of the class of functions in terms of that of
the simplest member of the class. Mostly we obtain the reduction formula by using integration by
parts. We can understand the method by the following examples.

Establish a reduction formula for ∫ 𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥.

Here the integrand is 𝑥 𝑛 𝑒 𝑎𝑥 . This is a general form of the function. In the reduction formula we seek
a relation of the given integral with another integral having the same form but involving 𝑛 − 1 or 𝑛 −
2 etc. And that can be accomplished by integrating the given integral by parts.
Let 𝐼𝑛 = ∫ 𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥
Integrating by parts, we get
𝑥 𝑛 𝑒 𝑎𝑥 𝑥 𝑛−1 𝑒 𝑎𝑥
𝐼𝑛 = −∫𝑛 𝑑𝑥
𝑎 𝑎

20 LOVELY PROFESSIONAL UNIVERSITY

Notes Calculus

𝑥 𝑛 𝑒 𝑎𝑥 𝑛
𝐼𝑛 = − 𝐼𝑛−1
𝑎 𝑎
Here you can see that 𝐼𝑛−1 is having the same type as that of given integral except that the ′𝑛′ has
reduced to ′𝑛 − 1′ . Hence you can see the justification for the name ‘reduction formula’ also.
Now if we need to calculate ∫ 𝑥 3 𝑒 𝑎𝑥 𝑑𝑥, we know what to do!
𝑥 𝑛 𝑒 𝑎𝑥 𝑛
In the reduction formula 𝐼𝑛 = − 𝐼𝑛−1 we will get 𝐼3 in terms of 𝐼2 , 𝐼2 in terms of 𝐼1 and 𝐼1 in
𝑎 𝑎
terms of 𝐼0 . The last integral in the recursion 𝐼0 is so easy to calculate. Then backward substitution
leads to the required integral.
Let’s see how it goes.
𝑥 3 𝑒 𝑎𝑥 3
𝐼3 = − 𝐼2 -----(1)
𝑎 𝑎
𝑥 2 𝑒 𝑎𝑥 2
𝐼2 = − 𝐼1 -----(2)
𝑎 𝑎
𝑥 1 𝑒 𝑎𝑥 1
𝐼1 = − 𝐼0 -----(3)
𝑎 𝑎

Solving the right side of (3).

𝑥1 𝑒 𝑎𝑥 1 𝑒 𝑎𝑥
𝐼1 = −
𝑎 𝑎 𝑎
𝑥 2 𝑒 𝑎𝑥 2 𝑥 1 𝑒 𝑎𝑥 1 𝑒 𝑎𝑥
𝐼2 = − ( − )
𝑎 𝑎 𝑎 𝑎 𝑎

𝑥 3 𝑒 𝑎𝑥 3 𝑥 2 𝑒 𝑎𝑥 2 𝑥1 𝑒 𝑎𝑥 1 𝑒 𝑎𝑥
𝐼3 = − ( − ( − ))
𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎

𝑥 3 𝑒 𝑎𝑥 3𝑥 2 𝑒 𝑎𝑥 6𝑥𝑒 𝑎𝑥 𝑒 𝑎𝑥
= − 2
+ 3
−6 4
𝑎 𝑎 𝑎 𝑎
which is the required solution.

Establish a reduction formula for ∫ 𝑥 𝑚 sin𝑛𝑥 𝑑𝑥.

In this integral, there are two parameters 𝑚 and 𝑛. In order to have a reduction formula, let us
integrate by parts taking 𝑥 𝑚 as the first function and sin 𝑛𝑥 as the second function.
𝑥 𝑚 𝑐𝑜𝑠𝑛𝑥 𝑚
∫ 𝑥 𝑚 sin𝑛𝑥 𝑑𝑥 = − + ∫ 𝑥 𝑚−1 cos 𝑛𝑥𝑑𝑥
𝑛 𝑛
Again integrating by parts,
𝑥 𝑚 𝑐𝑜𝑠𝑛𝑥 𝑚𝑥 𝑚−1 𝑠𝑖𝑛𝑛𝑥 𝑚(𝑚 − 1)
∫ 𝑥 𝑚 sin𝑛𝑥 𝑑𝑥 = − + − ∫ 𝑥 𝑚−2 sin 𝑛𝑥𝑑𝑥
𝑛 𝑛2 𝑛2
Therefore here we got the relation in two integrals of the same type but the one on the right side is of
lower degree. The left side integral can be written as 𝐼𝑚,𝑛 and the right-side integral which is reduced
version can be written as 𝐼𝑚−2,𝑛 .

∞
Establish a reduction formula for ∫ 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 and deduce that ∫0 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 = 𝑛! where 𝑛 is
any natural number.

∫ 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 = −𝑥 𝑛 𝑒 −𝑥 + 𝑛∫ 𝑥 𝑛−1 𝑒 −𝑥 𝑑𝑥
Or
𝐼𝑛 = −𝑥 𝑛 𝑒 −𝑥 + 𝑛𝐼𝑛−1 is the required reduction formula. Now for the deduction, consider the
integral,
𝑡 𝑡
∫ 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 = |−𝑥 𝑛 𝑒 −𝑥 |𝑡0 + 𝑛 ∫ 𝑥 𝑛−1 𝑒 −𝑥 𝑑𝑥
0 0
𝑡
= −𝑡 𝑛 𝑒 −𝑡 + 𝑛 ∫ 𝑥 𝑛−1 𝑒 −𝑥 𝑑𝑥
0

LOVELY PROFESSIONAL UNIVERSITY 21

 Unit 03: Integration by using reduction formula
Notes

The first term on the right-hand side tends to zero as 𝑡 → ∞.

∞ ∞
Therefore, ∫0 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 = ∫0 𝑥 𝑛−1 𝑒 −𝑥 𝑑𝑥

Or 𝐼𝑛 = 𝑛𝐼𝑛−1
𝐼𝑛−1 = (𝑛 − 1)𝐼𝑛−2
𝐼𝑛−2 = (𝑛 − 2)𝐼𝑛−3
𝐼𝑛−3 = (𝑛 − 3)𝐼𝑛−4
.
.
.
𝐼1 = 1. 𝐼0 = 1 (The integral 𝐼0 can easily be calculated to be 1.)
By back substitution we get
𝐼𝑛 = 𝑛(𝑛 − 1)(𝑛 − 2) … 3.2.1 = 𝑛!

𝑥𝑛
Establish the reduction formula for ∫ (𝑙𝑜𝑔𝑥)𝑚
𝑑𝑥

In this integral, there are two parameters, and the idea is to get a relation of the given integral with
another integral of same type but reduced parameter(s) any one or both.
To integrate the given function, let us write the integrand in the following way:
1 1
𝐼𝑚,𝑛 = ∫ 𝑥 𝑛+1 [ . ] 𝑑𝑥
(𝑙𝑜𝑔𝑥)𝑚 𝑥
Now the integration by parts can be applied taking the term in bracket as the second function.
Therefore,
(log 𝑥)−𝑚+1 (log 𝑥)−𝑚+1
𝐼𝑚,𝑛 = 𝑥 𝑛+1 − ∫ (𝑛 + 1)𝑥 𝑛 𝑑𝑥
−𝑚 + 1 −𝑚 + 1
(log 𝑥)−𝑚+1 𝑛 + 1 𝑥𝑛
𝐼𝑚,𝑛 = 𝑥 𝑛+1 + ∫ 𝑑𝑥
−𝑚 + 1 𝑚−1 (log 𝑥)𝑚−1
(log 𝑥)−𝑚+1 𝑛+1
Or 𝐼𝑚,𝑛 = 𝑥 𝑛+1 + 𝐼 is the required reduction formula.
−𝑚+1 𝑚−1 𝑚−1,𝑛

𝜋
Evaluate ∫02 sin𝑛 𝑥 𝑑𝑥 where 𝑛 is a positive integer.

The reduction formula for the integral is given as,

𝑛−1
𝐼𝑛 = 𝐼
𝑛 𝑛−2
𝑛−3
𝐼𝑛−2 = 𝐼
𝑛 − 2 𝑛−4
𝑛−5
𝐼𝑛−4 = 𝐼
𝑛 − 4 𝑛−6
.
.
.
2
𝐼3 = 𝐼1 if 𝑛 is odd
3
1
𝐼2 = 𝐼0 if 𝑛 is even
2

Therefore it can be compiled as

22 LOVELY PROFESSIONAL UNIVERSITY

Notes Calculus

𝑛−1 𝑛−3 𝑛−5 2

. . … 𝐼1 if 𝑛 is odd
𝐼𝑛 = { 𝑛 𝑛 − 2 𝑛 − 4 3
𝑛−1 𝑛−3 𝑛−5 1
. . … 𝐼 if 𝑛 is even
𝑛 𝑛−2 𝑛−4 2 0
𝜋
Now 𝐼1 = ∫02 𝑠𝑖𝑛𝑥 𝑑𝑥 = 1 and
𝜋
2 𝜋
𝐼0 = ∫ 1 𝑑𝑥 =
0 2
Thus the above integral can finally be written as
𝑛−1 𝑛−3 𝑛−5 2
. . … if 𝑛 is odd
𝐼𝑛 = { 𝑛 𝑛−2 𝑛−4 3
𝑛−1 𝑛−3 𝑛−5 1 𝜋
. . … . if 𝑛 is even
𝑛 𝑛−2 𝑛−4 2 2

Obtain a reduction formula for ∫ sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥.

∫ sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥 = ∫ sin𝑝−1 𝑥(sin 𝑥 cos𝑞 𝑥)𝑑𝑥

cos𝑞+1 𝑥 cos𝑞+1 𝑥
= sin𝑝−1 𝑥 (− ) − ∫ (𝑝 − 1) sin𝑝−2 𝑥. (− ) cos 𝑥 𝑑𝑥
𝑞+1 𝑞+1
sin𝑝−1 x cos𝑞+1 𝑥 𝑝−1
= (− )+ ∫ sin𝑝−2 𝑥. cos𝑞 𝑥 cos2 𝑥 𝑑𝑥
𝑞+1 𝑞+1
Now here we can use the trigonometric identity 𝑐𝑜𝑠 2 𝑥 = 1 − sin2 𝑥 in the last term.
𝑝−1 𝑝−1
(1 + ) ∫ sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥 = −(sin𝑝−1 𝑥𝑐𝑜𝑠 𝑞+1 𝑥)/(𝑞 + 1) + ∫ sin𝑝−2 𝑥 cos𝑞 𝑥 𝑑𝑥
𝑞+1 𝑝+𝑞
𝑝−1
∫ sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥 = (− sin𝑝−1 𝑥 cos𝑞+1 𝑥)/(𝑝 + 𝑞) + ∫ sin𝑝−2 𝑥 cos𝑞 𝑥 𝑑𝑥
𝑝+𝑞
is the required reduction formula.

𝑥𝑛
Construct the reduction formula for ∫ 𝑑𝑥 where 𝑛 ∈ 𝑵
√𝑎𝑥 2 +𝑏𝑥+𝑐

Here the integrand is an irrational function and it can be re written by involving the derivative of the
term in the denominator as follows:
2𝑎𝑥 + 𝑏 − 𝑏 𝑛−1
𝑥𝑛 = 𝑥
2𝑎
1 (2𝑎𝑥 + 𝑏)𝑥 𝑛−1 𝑏 𝑥 𝑛−1
𝐼𝑛 = ∫ 𝑑𝑥 − ∫ 𝑑𝑥
2𝑎 √𝑎𝑥 2 + 𝑏𝑥 + 𝑐 2𝑎 √𝑎𝑥 2 + 𝑏𝑥 + 𝑐
1 𝑛−1 𝑏(𝑛 − 1) 𝑐(𝑛 − 1) 𝑏
𝐼𝑛 = 𝑥 √𝑎𝑥 2 + 𝑏𝑥 + 𝑐 − (𝑛 − 1) 𝐼𝑛 − 𝐼𝑛−1 − 𝐼𝑛−2 − 𝐼
𝑎 𝑎 𝑎 2𝑎 𝑛−1
1 𝑏(2𝑛−1) 𝑐(𝑛−1)
𝐼𝑛 = 𝑥 𝑛−1 √𝑎𝑥 2 + 𝑏𝑥 + 𝑐 − 𝐼𝑛−1 − 𝐼𝑛−2 is the required reduction formula.
𝑛𝑎 2𝑛𝑎 𝑛𝑎
𝑥3
Let us evaluate ∫ 𝑑𝑥. You can see that this integral is a particular case of the integral whose
√𝑥 2 −2𝑥+2
reduction formula is
𝑥𝑛 1 𝑏(2𝑛−1) 𝑥 𝑛−1 𝑐(𝑛−1) 𝑥 𝑛−2
∫ 𝑑𝑥 = 𝑥 𝑛−1 √𝑎𝑥 2 + 𝑏𝑥 + 𝑐 − ∫ 𝑑𝑥 − ∫ 𝑑𝑥
√𝑎𝑥 2 +𝑏𝑥+𝑐 𝑛𝑎 2𝑛𝑎 √𝑎𝑥 2 +𝑏𝑥+𝑐 𝑛𝑎 √𝑎𝑥 2 +𝑏𝑥+𝑐

The above reduction formula can be written as:

𝑥3 1 5 𝑥2 4 𝑥1
∫ 𝑑𝑥 = 𝑥 2 √𝑥 2 − 2𝑥 + 2 + ∫ 𝑑𝑥 − ∫ 𝑑𝑥 -----(1)
√𝑥 2 −2𝑥+2 3 3 √𝑥 2 −2𝑥+2 3 √𝑥 2 −2𝑥+2

𝑥2 1 3 𝑥1 1
Also ∫ 𝑑𝑥 = √𝑥 2 − 2𝑥 + 2 + ∫ 𝑑𝑥 − ∫ 𝑑𝑥 -----(2)
√𝑥 2 −2𝑥+2 2 2 √𝑥 2 −2𝑥+2 √𝑥 2 −2𝑥+2

Let us work on the two integrals on the right hand side of the equation (2).

LOVELY PROFESSIONAL UNIVERSITY 23

 Unit 03: Integration by using reduction formula
Notes

𝑥 1 2𝑥 −2 1 2
∫ 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 𝑑𝑥 = (𝑥 2 − 2𝑥 + 2)1/2 + sinh−1 (𝑥 − 1) -----(3)
√𝑥 2 −2𝑥+2 2 √𝑥 2 −2𝑥+2 2 √(𝑥−1)2 +1

1
∫ 𝑑𝑥 = sinh−1 (𝑥 − 1) -----(4)
√𝑥 2 −2𝑥+2

Substituting (3) and (4) in (2), we get

𝑥2 1 3 1 1
∫ 𝑑𝑥 = √𝑥 2 − 2𝑥 + 2 + (𝑥 2 − 2𝑥 + 2)2 + sinh−1 (𝑥 − 1)
√𝑥 2 − 2𝑥 + 2 2 2 2
And the back substitution in equation (1) will give
𝑥3
∫ 𝑑𝑥
√𝑥 2 − 2𝑥 + 2
1 5 1 3 1 1
= 𝑥 2 √𝑥 2 − 2𝑥 + 2 + ( √𝑥 2 − 2𝑥 + 2 + (𝑥 2 − 2𝑥 + 2)2 + sinh−1 (𝑥 − 1) )
3 3 2 2 2
4
− ((𝑥 2 − 2𝑥 + 2)1/2 + sinh−1 (𝑥 − 1))
3
(2𝑥 2 + 12) 1
= − sinh−1 (𝑥 − 1)
6 2

𝑠𝑖𝑛𝑛𝑥
Form a reduction formula for ∫ 𝑑𝑥
𝑠𝑖𝑛𝑥

The parameter 𝑛 can be seen in the numerator of the integrand. Here integration by parts may not
help in getting a reduced form of the same type of integral. So we can use trigonometric identity
𝐶+𝐷 𝐶−𝐷
(sin 𝐶 − sin 𝐷 = 2 cos sin ) as follows:
2 2

sin 𝑛𝑥 − sin(𝑛 − 2) 𝑥 = 2 cos(𝑛 − 1) 𝑥 sin 𝑥

sin 𝑛𝑥 = sin(𝑛 − 2) 𝑥 + 2 cos(𝑛 − 1) 𝑥 sin 𝑥
Now we can rewrite the given integral as
𝑠𝑖𝑛𝑛𝑥 sin(𝑛 − 2) 𝑥 + 2 cos(𝑛 − 1) 𝑥 sin 𝑥
∫ 𝑑𝑥 = ∫ 𝑑𝑥
𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑛𝑥 sin(𝑛 − 2) 𝑥
∫ 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 2 cos(𝑛 − 1) 𝑥 𝑑𝑥
𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑛𝑥 sin(𝑛 − 2) 𝑥 2 sin(𝑛 − 1) 𝑥
∫ 𝑑𝑥 = ∫ 𝑑𝑥 +
𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥 𝑛−1
which is the required reduction formula connecting the given integral with its reduced version.

3.2 Properties of definite integral

Before studying the properties of definite integral, let us recapitulate some basics about them.

Definite Integral
𝑏
The definite integral is an integral of the form ∫𝑎 𝑓(𝑥) 𝑑𝑥. This integral is read as the integral from a
to b of 𝑓(𝑥) 𝑑𝑥. The numbers a and b are said to be the limits of integration. For our problems, a is less
than b. Definite Integrals are evaluated using the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus

Let 𝑓(𝑥) be a continuous function for 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝐹(𝑥) be an anti-derivative of 𝑓(𝑥). Then
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = [𝐹(𝑥)]|𝑏𝑎 = 𝐹(𝑏) − 𝐹(𝑎).
If 𝑓(𝑥) ≥ 0 for 𝑎 ≤ 𝑥 ≤ 𝑏. Then
𝑏 Area Between 𝑓(𝑥) and the
Definite Integral: ∫𝑎 𝑓(𝑥) 𝑑𝑥 =
𝑥 axis for 𝑎 ≤ 𝑥 ≤ 𝑏
The Second Fundamental Theorem of Calculus
If f is continuous on an open interval, I containing a, then for every x in the interval
𝑥
𝑑
[∫ 𝑓(𝑡) 𝑑𝑡 ] = 𝑓(𝑥)
𝑑𝑥 𝑎

24 LOVELY PROFESSIONAL UNIVERSITY

Notes Calculus

Chain Rule of Differentiation

𝑑
(𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥)
𝑑𝑥
Suppose we let 𝐹 = 𝑓(𝑔(𝑥)) and let 𝑢 = 𝑔(𝑥). Then 𝐹 = 𝑓(𝑢) and hence
𝑑𝐹 𝑑𝑢
= 𝑓 ′ (𝑢) and = 𝑔′ (𝑥)
𝑑𝑢 𝑑𝑥

For 𝐹 = 𝑓(𝑔(𝑥)) and 𝑢 = 𝑔(𝑥) we can use these ideas to rewrite the chain rule as follows:
𝑑𝐹 𝑑 𝑑𝑢 𝑑𝐹 𝑑𝑢
𝐹 ′ (𝑥) = = (𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥) = 𝑓 ′ (𝑢) =
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥
This gives another way to write the chain rule, which is as follows:

Chain Rule: Alternative Form

If we want to differentiate the composition𝐹(𝑥) = 𝑓(𝑔(𝑥)), we set 𝑢 = 𝑔(𝑥) and compute the
𝑑𝐹 𝑑𝐹 𝑑𝑢
following: 𝐹 ′ (𝑥) = = ⋅
𝑑𝑥 𝑑𝑢 𝑑𝑥

For example, suppose we want to differentiate𝐹(𝑥) = (𝑥 2 + 7)10

Then by taking, 𝑢 = 𝑥 2 + 7, we have 𝐹 = 𝑢10
𝑑𝐹 𝑑𝑢
Hence, = 10𝑢9 and = 2𝑥.
𝑑𝑢 𝑑𝑥

Therefore, for 𝐹(𝑥) = (𝑥 2 + 7)10

𝑑𝐹 𝑑𝐹 𝑑𝑢
𝐹 ′ (𝑥) = = ⋅ = 10𝑢9 (2𝑥) = 10(𝑥 2 + 7)9 (2) = 20𝑥(𝑥 2 + 7)9.
𝑑𝑥 𝑑𝑢 𝑑𝑥

This way of expressing the chain rule can be useful when using the Second Fundamental Theorem of
𝑔(𝑥) 𝑢
Calculus. Suppose 𝐹(𝑥) = ∫𝑎 𝑓(𝑡) 𝑑𝑡 , then taking 𝑢 = 𝑔(𝑥) gives 𝐹 = ∫𝑎 𝑓(𝑡) 𝑑𝑡
𝑑𝐹 𝑑𝑢 𝑑 𝑢 𝑑𝑢 𝑑𝑢
Then 𝐹 ′ (𝑥) = ⋅ = (∫𝑎 𝑓(𝑡) 𝑑𝑡 ) ⋅ = 𝑓(𝑢) ⋅ = 𝑓(𝑔(𝑥)) ⋅ 𝑔′ (𝑥)
𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑥

Properties
𝑏 𝑏
1. ∫𝑎 𝜙(𝑥)𝑑𝑥 = ∫𝑎 𝜙(𝑡)𝑑𝑡
Let ∫ 𝜙(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐1
and ∫ 𝜙(𝑡)𝑑𝑡 = 𝐹(𝑡) + 𝑐2
𝑏
Therefore, ∫𝑎 𝜙(𝑥)𝑑𝑥 = [𝐹(𝑥) + 𝑐1 ]𝑏𝑎 = 𝐹(𝑏) − 𝐹(𝑎)
𝑏
Similarly, ∫𝑎 𝜙(𝑡)𝑑𝑡 = 𝐹(𝑏) − 𝐹(𝑎)

This property explains the dummy nature of the variable of integration in a definite integral.
𝑏 𝑐 𝑏
2. ∫𝑎 𝜙(𝑥)𝑑𝑥 = ∫𝑎 𝜙(𝑥)𝑑𝑥 + ∫𝑐 𝜙(𝑥)
Let ∫ 𝜙(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐1
Then the RHS= [𝐹(𝑥) + 𝑐1 ]𝑐𝑎 + [𝐹(𝑥) + 𝑐1 ]𝑏𝑐
= 𝐹(𝑐) − 𝐹(𝑎) + 𝐹(𝑏) − 𝐹(𝑐)
= 𝐹(𝑏) − 𝐹(𝑎)
𝑏
= ∫ 𝜙(𝑥)𝑑𝑥 = 𝐿𝐻𝑆
𝑎
𝑏 𝑎
3. ∫𝑎 𝜙(𝑥)𝑑𝑥 = − ∫𝑏 𝜙(𝑥)𝑑𝑥
Let ∫ 𝜙(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐1
𝑎
Consider the RHS = − ∫𝑏 𝜙(𝑥)𝑑𝑥

= −[𝐹(𝑥) + 𝑐1 ]𝑎𝑏
= [𝐹(𝑎) + 𝑐1 − 𝐹(𝑏) − 𝑐1

LOVELY PROFESSIONAL UNIVERSITY 25

 Unit 03: Integration by using reduction formula
Notes

𝑏
= 𝐹(𝑏) − 𝐹(𝑎) = ∫ 𝜙(𝑥)𝑑𝑥 = 𝐿𝐻𝑆
𝑎
𝑎 𝑎
4. ∫0 𝜙(𝑥)𝑑𝑥 = ∫0 𝜙(𝑎 − 𝑥)𝑑𝑥
Letting 𝑎 − 𝑥 = 𝑡
or 𝑥 = 𝑎 − 𝑡
or 𝑑𝑥 = −𝑑𝑡
𝑎 0 𝑎 𝑎
Now the RHS= ∫0 𝜙(𝑎 − 𝑥)𝑑𝑥 = − ∫𝑎 𝜙(𝑡)𝑑𝑡 = ∫0 𝜙(𝑡)𝑑𝑡 = ∫0 𝜙(𝑥)𝑑𝑥
𝑎
2𝑎 2 ∫0 𝜙(𝑥)𝑑𝑥 if 𝜙(2𝑎 − 𝑥) = 𝜙(𝑥)
5. ∫0 𝜙(𝑥)𝑑𝑥 = {
0 if 𝜙(2𝑎 − 𝑥) = −𝜙(𝑥)
2𝑎
Consider ∫𝑎 𝜙(𝑥)𝑑𝑥

Put 2𝑎 − 𝑥 = 𝑡 This implies −𝑑𝑥 = 𝑑𝑡. Therefore,

2𝑎 0
∫ 𝜙(𝑥)𝑑𝑥 = − ∫ 𝜙(2𝑎 − 𝑡)𝑑𝑡
𝑎 𝑎
𝑎
= ∫ 𝜙(2𝑎 − 𝑡)𝑑𝑡
0
𝑎
= ∫ 𝜙(2𝑎 − 𝑥)𝑑𝑥
0
2𝑎 𝑎 2𝑎
Now ∫0 𝜙(𝑥)𝑑𝑥 = ∫0 𝜙(𝑥)𝑑𝑥 + ∫𝑎 𝜙(𝑥)𝑑𝑥
𝑎 𝑎
= ∫ 𝜙(𝑥)𝑑𝑥 + ∫ 𝜙(2𝑎 − 𝑥)𝑑𝑥
0 0
𝑎
The second integrand on the RHS can be 𝜙(𝑥) or −𝜙(𝑥) and then accordingly it will yield 2 ∫0 𝜙(𝑥)𝑑𝑥
or a zero. Hence the property.

𝑎 0 if 𝜙(−𝑥) = −𝜙(𝑥)
6. ∫−𝑎 𝜙(𝑥)𝑑𝑥 = { 𝑎
2 ∫0 𝜙(𝑥)𝑑𝑥 if 𝜙(−𝑥) = 𝜙(𝑥)
0
Consider ∫−𝑎 𝜙(𝑥)𝑑𝑥

Put 𝑥 = −𝑡, 𝑑𝑥 = −𝑑𝑡

0 0 𝑎
Then ∫−𝑎 𝜙(𝑥)𝑑𝑥 = − ∫𝑎 𝜙(−𝑡)𝑑𝑡 = ∫0 𝜙(−𝑥)𝑑𝑥
𝑎 0 𝑎
Therefore, ∫−𝑎 𝜙(𝑥)𝑑𝑥 = ∫−𝑎 𝜙(𝑥)𝑑𝑥 + ∫0 𝜙(𝑥)𝑑𝑥
0 if 𝜙(−𝑥) = −𝜙(𝑥)
𝑎
={
2 ∫ 𝜙(𝑥)𝑑𝑥 if 𝜙(−𝑥) = 𝜙(𝑥)
0

Now let us evaluate a few integrals using the properties.

𝜋/2
Evaluate ∫0 log(1 + 𝑡𝑎𝑛𝜃) 𝑑𝜃
𝜋
Let 𝐼 = ∫04 log(1 + 𝑡𝑎𝑛𝜃) 𝑑𝜃
𝜋
𝜋
= ∫04 log (1 + tan ( − 𝜃)) 𝑑𝜃
4
𝜋
4 1 − 𝑡𝑎𝑛𝜃
= ∫ log (1 + ) 𝑑𝜃
0 1 + 𝑡𝑎𝑛𝜃
𝜋
4 2
= ∫ log ( ) 𝑑𝜃
0 1 + 𝑡𝑎𝑛𝜃

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𝜋
𝜋 4
= log 2 − ∫ log(1 + 𝑡𝑎𝑛𝜃) 𝑑𝜃
4 0
𝜋
Or 𝐼 = log 2
8

𝜋
Evaluate ∫0 𝑥 sin6 𝑥 cos4 𝑥 𝑑𝑥

𝜋
Let 𝐼 = ∫0 𝑥 sin6 𝑥 cos4 𝑥 𝑑𝑥
𝜋
= ∫ (𝜋 − 𝑥) sin6 (𝜋 − 𝑥) cos4 (𝜋 − 𝑥)𝑑𝑥
0
𝜋 𝜋
= ∫ (𝜋) sin6 (𝑥) cos4 (𝑥)𝑑𝑥 − ∫ (𝑥) sin6 (𝑥) cos4 (𝑥)𝑑𝑥
0 0
𝜋/2
2𝐼 = 2𝜋 ∫ sin6 𝑥 cos4 𝑥 𝑑𝑥
0

2𝜋 5.3.1.3.1 𝜋
𝐼=
2 10.8.6.4.2 2
3𝜋 2
𝐼=
512

Summary
In this chapter we have seen one more technique of solving the integral by writing a recursion
formula. Many general integrals can be solved by this method including their particular cases. We
have also seen the proofs of the properties of the definite integral.

Key words
Definite integral, reduction formula, properties of definite integral

Self-Assessment
3
1. ∫2 𝑥 3 𝑑𝑥 is equal to
(a) 65 (b) 65/4 (c) ¼ (d) 63/4

𝜋
2. If 𝑚 ≠ 𝑛, then ∫0 cos 𝑚𝑥 cos 𝑛𝑥 𝑑𝑥 is
𝜋
(a) 0 (b) (c) 𝜋 (d) 2𝜋
2

𝜋
3. ∫02 sin5 𝑥 𝑑𝑥 is
(a) 5/15 (b) 6/15 (c) 7/15 (d) 8/15
𝜋
4. ∫02 sin6 𝑥 𝑑𝑥 is
(a) 5/32 (b) 5𝜋/32 (c) 5/16 (d) 5𝜋/16

𝜋
5. ∫02 cos7 𝑥 𝑑𝑥 is
(a) 16/35 (b) 6/15 (c) 17/15 (d) 8/35

𝜋
6. ∫02 cos8 𝑥 𝑑𝑥 is
(a) 5/32 (b) 35𝜋/256 (c) 35/256 (d) 5𝜋/16

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 Unit 03: Integration by using reduction formula
Notes

𝜋
7. ∫04 cos6 2𝑥 𝑑𝑥 is
(a) 5/64 (b) 5𝜋/64 (c) 35/256 (d) 𝜋/16

3
1
8. ∫0 𝑥 2 (1 − 𝑥 2 )2 𝑑𝑥 is
𝜋 𝜋 𝜋 𝜋
(a) (b) (c) (d)
32 16 8 4

𝜋/2 √𝑠𝑖𝑛𝑥
9. ∫0 𝑑𝑥 is
√𝑠𝑖𝑛𝑥+√𝑐𝑜𝑠𝑥
𝜋 𝜋 𝜋
(a) (b) (c) (d) None of these
2 4 8

𝜋
10. ∫0 𝑥𝑠𝑖𝑛6 𝑥𝑐𝑜𝑠 4 𝑥 𝑑𝑥 is
3𝜋2 3𝜋2 𝜋2
(a) 3𝜋 2 (b) (c) (d)
51 512 512

𝜋
11. ∫02 log sin 𝑥 𝑑𝑥 is
𝜋 𝜋 3𝜋 𝜋
(a) − log 2 (b) log 2 (c) − log 2 (d) −
2 2 2 2

𝜋
12. ∫0 log (1 + cos 𝑥)𝑑𝑥 is
(a) −𝜋 log 2 (b) 𝜋 log 1/2 (c) Both (a) and (b) (d) None of these

𝜋/2
13. ∫0 sin5 𝑥 cos6 𝑥 𝑑𝑥 is
(a) 8/693 (b) 8/69 (c) 8/6 (d) None of these

𝜋/2
14. ∫0 sin6 𝑥 cos8 𝑥 𝑑𝑥 is
5 5𝜋 𝜋
(a) (b) (c) (d) None of these
4096 4096 4096

𝜋/2
15. ∫0 cos 2𝑥 cos3 𝑥 𝑑𝑥 is
8 2 2 5
(a) (b) (c) (d)
15 15 5 2

Answers:

1 b 2 a 3 d 4 b 5 a

6 b 7 b 8 a 9 b 10 c

11 a 12 c 13 a 14 b 15 c

Review Questions
1. Construct the reduction formula for ∫ sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥 where 𝑝, 𝑞 are positive integers.
𝜋/2
2. Evaluate the definite integral ∫0 sin𝑝 𝑥 cos𝑞 𝑥 𝑑𝑥 where 𝑝, 𝑞 are positive integers.
𝜋
3. Using the properties prove that ∫0 (𝑥 sin 𝑥)/(1 + cos2 𝑥) 𝑑𝑥 = 𝜋 2 /4.
𝜋/2 1
4. Using the properties prove that ∫0 sin2 𝑥/(sin 𝑥 + 𝑐𝑜𝑠 𝑥) 𝑑𝑥 = log(√2 + 1)
√2
𝜋/2
5. Evaluate ∫0 𝑥 cot 𝑥 𝑑𝑥 using the properties of definite integral.

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Notes Calculus

Further/Suggested Readings
George B. Thomas Jr., Joel Hass, Christopher Heil & Maurice D. Weir (2018).
Thomas’ Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens & Stephan Davis (2016). Calculus (10th edition).
Wiley India.

https://www.intmath.com/integration/4-definite-integral.php
https://www.math24.net/properties-definite-integrals
https://www.intmath.com/methods-integration/10-integration-reduction-
formulae.php

LOVELY PROFESSIONAL UNIVERSITY 29

Dr. Pratiksha, Lovely Professional University Unit 04: Limit of a Real Valued Function

Notes

Unit 04: Limit of a Real Valued Function

CONTENTS
Objectives
Introduction
4.1 Real valued function
4.2 Epsilon delta definition of the limit
4.3 Some results on limits
4.4 Limit at infinity and infinite limits
Summary
Key words
Review Questions
Further/Suggested Readings

Objectives
Students will

• learn about the definition of real valued function

• learn about the limit of the function
• understand the epsilon delta definition of limit
• learn about infinite limit and limit at infinity

Introduction
In this unit we will understand one of the most crucial and fundamental concept of calculus. But
before that let us have an idea about the function. You can consider a function as a kind of rule
where you give in some input and get a specific output. The input is decided as everything that
keeps the function well defined. The technical name for such input is the domain. Let us see what a
function is! First of all there must be two non-empty sets 𝐴 and 𝐵, then a function 𝑓 from 𝐴 to 𝐵 is
denoted as 𝑓: 𝐴 → 𝐵 and is defined as a function if for all the values in set 𝐴, there corresponds a
unique value in set B. Set A is called the domain, B is called the codomain and 𝑓(𝐴) is called the
range.

4.1 Real valued function

The function can be expressed as a set, a formula, a table or as a graph. If the range of the function
is a set of real numbers, then it is called a real valued function.
For example: {(2,4), (3,9), (4,16), (5,25)} is a function where domain is the set {2,3,4,5} and range is
the set {4,9,16,25}. Clearly, the domain and range are discrete in this case and are subsets of real
numbers. In other form 𝑦 = 𝑥 2 is a function. The input can be any real number as any real
number will keep the function well defined and with any real input, the output is always going
to be a non-negative real number. Therefore the range is set of non-negative real number. The
same function can be represented in the form of the following graph:

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 Calculus

Notes

We can show the first representation of function in the form of a table also, which is as follows:

x y

2 4

3 9

4 16

5 25

Of course here the domain and range are discrete and finite subsets of real numbers.
𝑥 2 −9
Now let us consider a function 𝑓(𝑥) = . It is quite clear that the function is not well defined at
𝑥−3
𝑥 = 3, and is good for all other real numbers. Therefore the domain is set of all real numbers except
3. If you input these values of the domain, the function can give any real number as the output,
except 6.
Now we know what happens at 𝑥 = 3, can we just observe the behaviour of the function as 𝑥 goes
closer to 3 from all possible directions? See the following table and see the pattern of 𝑓(𝑥) as 𝑥 goes
close to 3 from left as well as from the right direction.

x f(x)

2.9 5.9

2.99 5.99

2.999 5.999

3.01 6.01

3.001 6.001

3.0001 6.0001

So, you can observe that as 𝑥 is approaching to 3, the 𝑓(𝑥) is approaching to 6. Then we say that the
limit of the function exists at 𝑥 = 3.

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 Unit 04: Limit of a Real Valued Function

Notes
If the function does not approach to the same value from both directions, then we say that the limit
of the function does not exist.
Here it must be mentioned that the word ‘close’ that has been used to define the concept of limit is
not a crisp word. It can mean something to me and entirely different thing to you. This closeness
has to be quantized in order to have a crisp definition of limit and that has been achieved through
the epsilon delta definition.

4.2 Epsilon delta definition of the limit

A function is said to tend to limit 𝑙 as 𝑥 tends to 𝑐, if ∀𝜖 > 0 however small, ∃ 𝛿 > 0, such that
|𝑓(𝑥) − 𝑙| < 𝜖 whenever 0 < |𝑥 − 𝑐| < 𝛿
or
𝑓(𝑥) ∈ (𝑙 − 𝜖, 𝑙 + 𝜖) ∀ 𝑥 ∈ (𝑐 − 𝛿, 𝑐) 𝑈 (𝑐, 𝑐 + 𝛿)
The quantity 𝜖 is how close you would like 𝑓(𝑥) to be to its limit 𝑙; the quantity 𝛿
is how close you have to choose 𝑥 to 𝑐 to achieve this.

𝑙𝑖𝑚
1. Prove that 𝑥→−14𝑥 + 1 = −3

To prove that 𝑥→𝑐

𝑙𝑖𝑚𝑓(𝑥) = 𝑙 you can assume that someone has given you some small positive value of

𝜖 and you need to find a positive value of 𝛿 for which |𝑓(𝑥) − 𝑙| < 𝜖 whenever 0 < |𝑥 − 𝑐| < 𝛿
holds. This 𝛿 surely depends on 𝜖.
Here we want to find the 𝛿 such that whenever |𝑥 + 1| < 𝛿, |4𝑥 + 1 + 3| < 𝜖 for a predefined 𝜖.
If we work out on the epsilon inequality, we can see |4(𝑥 + 1)| < 𝜖 . That is |𝑥 + 1| < 𝜖/4. Now we can do
𝜖
a smart work here. If we consider 𝜖/4 as 𝛿, we are done. Thus for a given 𝜖 and 𝛿 = , we have |4𝑥 + 1 +
4
3| < 𝜖 whenever |𝑥 + 1| < 𝜖/4.

𝑙𝑖𝑚 1
2. Prove that 𝑥→0 𝑥𝑠𝑖𝑛 ( ) = 0
𝑥

Let 𝜖 > 0 be given. Then we would like to find a 𝛿 > 0, such that
|𝑓(𝑥) − 0| < 𝜖 whenever |𝑥 − 0| < 𝛿
Now
1
|𝑓(𝑥) − 0| = |𝑥𝑠𝑖𝑛 ( )|
𝑥
1
= |𝑥| |sin ( )|
𝑥
≤ |𝑥|
Now choosing 𝛿 = 𝜖 we can see that
|𝑓(𝑥) − 0| < 𝜖 whenever |𝑥| < 𝜖
𝑙𝑖𝑚 1
∴ 𝑥→0 𝑥 sin ( ) = 0
𝑥

4.3 Some results on limits

The precise definition of the limit is not so easy to use, and we won't use it very often in this course.
Instead, there are a number of properties that limits have, which allow you to compute them
without having to use the epsilon delta definition. Let us see some of the properties of limit in the
form of following results.
Let 𝑓 and 𝑔 are two functions such that 𝑥→𝑎
𝑙𝑖𝑚 𝑓(𝑥) = 𝑙 and 𝑙𝑖𝑚
𝑥→𝑎 𝑔(𝑥) = 𝑚. Then,
𝑙𝑖𝑚 𝑙𝑖𝑚
𝑥→𝑎 (𝑓 + 𝑔)(𝑥) = 𝑥→𝑎 (𝑓(𝑥) + 𝑔(𝑥)) = 𝑙 + 𝑚

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Notes Here you can see that the limit of sum of two functions is the sum of their limits. In simple words, you can see
that as 𝑓(𝑥) is getting close to 𝑙 and 𝑔(𝑥) is getting close to 𝑚 as 𝑥 → 𝑎, then 𝑓(𝑥) + 𝑔(𝑥) will go close to
𝑙 + 𝑚 only. Though it can be proved by definition, but here we will resort to common sense only. Similarly
we can see some more properties:
𝑙𝑖𝑚 𝑙𝑖𝑚
𝑥→𝑎 (𝑓 − 𝑔)(𝑥) = 𝑥→𝑎 (𝑓(𝑥) − 𝑔(𝑥)) = 𝑙 − 𝑚
The limit of difference of two functions is the difference in the respective limit of the individual functions.
𝑙𝑖𝑚 𝑙𝑖𝑚
𝑥→𝑎 (𝑓𝑔)(𝑥) = 𝑥→𝑎 𝑓(𝑥)𝑔(𝑥) = 𝑙 𝑚
The limit of the product of two functions is the product of the respective limits of the individual functions.
𝑙𝑖𝑚 (𝑓/𝑔)(𝑥) = 𝑙𝑖𝑚 𝑓(𝑥)/𝑔(𝑥) = 𝑙 /𝑚 (𝑚 ≠ 0)
𝑥→𝑎 𝑥→𝑎

The limit of the quotient of two functions is the quotient of the limits of respective functions.

lim 1−cos 𝑥
3. Evaluate 𝑥→0 𝑥2

𝑥 𝑥 2
1 − cos 𝑥 2 sin2 1 sin 1
lim
= lim 2= lim
( 2
𝑥→0
𝑥2 𝑥→0
𝑥2 𝑥→0
2 𝑥 ) =2
2

lim 𝑒 𝑡𝑎𝑛𝑥 −𝑒 𝑥
4. Evaluate 𝑥→0 tan 𝑥−𝑥

lim 𝑒 𝑡𝑎𝑛𝑥 −𝑒 𝑥 lim 𝑒 𝑥(𝑒 𝑡𝑎𝑛𝑥−𝑥 −1)

𝑥→0 = 𝑥→0 = 𝑒0. 1 = 1
tan 𝑥−𝑥 tan 𝑥−𝑥

𝑙𝑖𝑚 ( 1𝑥 +2𝑥 +⋯+𝑛𝑥 𝑎/𝑥

5. Evaluate 𝑥→0 )
𝑛

lim 𝑥 lim
Note that 𝑥→0 (1 + 𝑓(𝑥)) = 𝑒 𝑥→0 𝑓(𝑥).𝑥
Now we can modify the given function, so that it takes the form of the left hand side of the above result.

𝑙𝑖𝑚 1𝑥 +2𝑥 +⋯+𝑛𝑥 𝑎/𝑥 𝑙𝑖𝑚 1𝑥 +2𝑥 +⋯+𝑛𝑥 𝑎/𝑥

Then, 𝑥→0 ( ) = 𝑥→0 (1 + − 1)
𝑛 𝑛

𝑙𝑖𝑚 1𝑥 +2𝑥 +⋯+𝑛𝑥 −𝑛 𝑎/𝑥

= 𝑥→0 (1 + )
𝑛
𝑎/𝑥
𝑙𝑖𝑚 1𝑥 −1+2𝑥 −1+⋯+𝑛𝑥 −1
( )
= 𝑒 𝑥→0 𝑛

𝑎
𝑙𝑖𝑚 𝑥𝑙𝑜𝑔1+𝑥𝑙𝑜𝑔2+⋯+𝑥𝑙𝑜𝑔𝑛 𝑥
( )
= 𝑒 𝑥→0 𝑛
𝑎
𝑙𝑖𝑚 𝑥𝑙𝑜𝑔 𝑛! 𝑥
𝑥→0 ( )
=𝑒 𝑛

𝑙𝑖𝑚 𝑎𝑥𝑙𝑜𝑔 𝑛!
= 𝑒 𝑥→0 𝑥𝑛

𝑎
= 𝑛! 𝑛

4.4 Limit at infinity and infinite limits

Here we will learn when do we say that a function is approaching to infinity as 𝑥 is approaching to
any number, as 𝑥 → ∞ , when a function has a finite limit and what is the behaviour of the function
as 𝑥 approaches to positive/negative infinity.

𝑙𝑖𝑚 𝑓(𝑥) = ∞
𝑥→𝑐

A function 𝑓 is said to tend to ∞ as 𝑥 tends to 𝑐, if for any 𝐺 > 0, however large, there corresponds a 𝛿 > 0
such that

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 Unit 04: Limit of a Real Valued Function

Notes
∀ 𝑥 ∈ (𝑐 − 𝛿)𝑈(𝑐, 𝑐 + 𝛿) , 𝑓(𝑥) > 𝐺

𝑙𝑖𝑚 𝑓(𝑥) = −∞
𝑥→𝑐

A function 𝑓 is said to tend to −∞ as 𝑥 tends to 𝑐, if for any 𝐺 > 0, however large, there corresponds a 𝛿 > 0
such that
∀ 𝑥 ∈ (𝑐 − 𝛿)𝑈(𝑐, 𝑐 + 𝛿) , 𝑓(𝑥) < −𝐺

𝑙𝑖𝑚 𝑓(𝑥) = 𝑙
𝑥→∞

A function 𝑓 is said to tend to 𝑙 as 𝑥 tends to ∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such that
|𝑓(𝑥) − 𝑙| < 𝜖 , ∀ 𝑥 > 𝐺
𝑙𝑖𝑚
𝑥→−∞ 𝑓(𝑥) = 𝑙
A function 𝑓 is said to tend to 𝑙 as 𝑥 tends to −∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such that
|𝑓(𝑥) − 𝑙| < 𝜖 , ∀ 𝑥 < −𝐺

𝑙𝑖𝑚 𝑓(𝑥) = ∞
𝑥→∞

A function 𝑓 is said to tend to ∞ as 𝑥 tends to ∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such that
𝑓(𝑥) > 𝜖 , ∀ 𝑥 > 𝐺

𝑙𝑖𝑚 𝑓(𝑥) = ∞
𝑥→−∞

A function 𝑓 is said to tend to ∞ as 𝑥 tends to −∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such
that
𝑓(𝑥) > 𝜖 , ∀ 𝑥 < −𝐺

𝑙𝑖𝑚
𝑥→∞ 𝑓(𝑥) = −∞
A function 𝑓 is said to tend to −∞ as 𝑥 tends to ∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such
that
𝑓(𝑥) < 𝜖 , ∀ 𝑥 > 𝐺

𝑙𝑖𝑚 𝑓(𝑥) = −∞
𝑥→−∞

A function 𝑓 is said to tend to −∞ as 𝑥 tends to −∞, if for any given 𝜖 > 0, there corresponds a 𝐺 > 0 such
that
𝑓(𝑥) < 𝜖 , ∀ 𝑥 < −𝐺

Let us see some problems now!

𝑙𝑖𝑚 1 𝑙𝑖𝑚 1 𝑙𝑖𝑚 1

1. Prove that 𝑥→0+ 𝑥 =∞, 𝑥→0− 𝑥 = −∞ and 𝑥→0 does not exist.
𝑥

1
Let = 𝑦
𝑥

Consider the case when 𝑥 > 0. This implies that 𝑦 > 0.

Let 𝐺 > 0 be any number, then
1 1
> 𝐺 if 0 < 𝑥 <
𝑥 𝐺

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Notes 𝑙𝑖𝑚 1
This implies that 𝑥→0+ 𝑥 =∞

Now consider the case when 𝑥 < 0. This implies that 𝑦 < 0.
Let 𝐺 > 0 be any number, then
1 1
< −𝐺 if − < 𝑥 < 0
𝑥 𝐺

𝑙𝑖𝑚 1
This implies 𝑥→0− = −∞
𝑥

𝑙𝑖𝑚 1
Therefore the left hand limit is different from the right hand limit as 𝑥 is approaching to zero and 𝑥→0 𝑥 does
not exist.

2𝑥+1 𝑙𝑖𝑚 𝑙𝑖𝑚

2. For 𝑓(𝑥) = show that 𝑥→∞ 𝑓(𝑥) = 2 and 𝑥→−∞ 𝑓(𝑥) = 2
𝑥−3

Let 𝜖 > 0 be given

2𝑥+1 7 7
Now | − 2| = |𝑥−3| < 𝜖 for 𝑥 > + 3
𝑥−3 𝜖

∴ 𝑙𝑖𝑚
𝑥→∞𝑓(𝑥) =2
2𝑥+1 7 7
Again | − 2| = |𝑥−3| < 𝜖 for 𝑥 < − + 3
𝑥−3 𝜖

∴ 𝑙𝑖𝑚
𝑥→−∞𝑓(𝑥) =2

If you look at the definitions of limits at infinity, you can find a positive number 𝐺, in both cases, which
fulfills the required criterion.

𝑙𝑖𝑚
3. Evaluate 𝑥→∞ (√𝑥 2 + 𝑥 + 1 − √𝑥 2 + 1 )

𝑙𝑖𝑚 (√𝑥 2 + 𝑥 + 1 − √𝑥 2 + 1 )
𝑥→∞

𝑙𝑖𝑚
𝑥
= 𝑥→∞
√𝑥 2 + 𝑥 + 1 + √𝑥 2 + 1
𝑙𝑖𝑚
1 1
= 𝑥→∞ =
2
√ 12 + 1 + 1 + √ 12 + 1
𝑥 𝑥 𝑥

Prove by epsilon delta definition that 𝑙𝑖𝑚(3𝑥 − 4) = 2.

𝑥→2

L’Hopital Rule

We have got a pretty good idea that for a quotient of two functions, such that the individual limit of
numerator is non-zero and of denominator is zero, then the overall limit of the quotient function
does not exist. In case the individual limit of numerator is zero and that of the denominator is non-
zero, then the overall limit of the quotient function is zero. The third case needs a special attention.
If the individual limit of both numerator as well as the denominator is zero, then this is called one
of the indeterminate forms and there are chances to get its value by using L’Hopital‘s rule.
sin 𝑥
Consider 𝑙𝑖𝑚 . Here the individual limit of the numerator is zero and of the denominator is also
𝑥→0 𝑥
0
zero. This is form. To evaluate this, we will have the following rule (for derivation you can see the
0
link 4 in the last section of the chapter).
Suppose 𝑓 and 𝑔 are differentiable functions over an open interval containing 𝑎, except possibly at
𝑎.

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 Unit 04: Limit of a Real Valued Function

Notes
𝑙𝑖𝑚
If 𝑥→a 𝑙𝑖𝑚
𝑓(𝑥) = 0 and 𝑥→a 𝑙𝑖𝑚
𝑔(𝑥) = 0 then 𝑥→a 𝑙𝑖𝑚
𝑓(𝑥)/g(x) = 𝑥→a 𝑓′(𝑥)/g′(x) assuming the limit on the
right exists or is ∞ or −∞. This result also holds if we are considering one-sided limits, or if 𝑎 = ∞
or 𝑎 = −∞.
sin 𝑥 0
Therefore using the above result for 𝑙𝑖𝑚 as it is a form. We can write the given limit as
𝑥→0 𝑥 0
cos 𝑥
𝑙𝑖𝑚 . And the limit can be evaluated to be one.
𝑥→0 1
0 ∞
There are mainly seven indeterminate forms and we try to convert them in or in form first if
0 ∞
they are not so. And the by the above mentioned formula the limit can be evaluated. Let us see one
more question for better clarity.

Let us evaluate 𝑙𝑖𝑚(𝑥log 𝑥) . Clearly it is 0 ⋅ ∞ form. So first of all we will rewrite the given
𝑥→0
function in the desirable (0/0 or ∞/∞) form.

log 𝑥
𝑙𝑖𝑚(𝑥log 𝑥). = 𝑙𝑖𝑚
𝑥→0 𝑥→01/𝑥
1
= 𝑙𝑖𝑚 𝑥
𝑥→0 1
− 2
𝑥
= 𝑙𝑖𝑚(−𝑥)
𝑥→0

=0

Now let us find the 𝑙𝑖𝑚(cot 𝑥)1/log 𝑥 . Clearly this is an indeterminate form of type ∞0 . First
𝑥→0
of all we will rewrite the given function into the required form. Here you can see the
function as a power of another function. So logarithm can simplify the system.

1
log 𝑦 = log(cot 𝑥)
log 𝑥
log cot 𝑥
⇒ 𝑙𝑖𝑚log 𝑦 = 𝑙𝑖𝑚
𝑥→0 𝑥→0 log 𝑥
−cosec 2 𝑥/ cot 𝑥
= 𝑙𝑖𝑚
𝑥→0 1
𝑥
−𝑥 1
= 𝑙𝑖𝑚 ⋅
𝑥→0 sin 𝑥 cos 𝑥

= −1

⇒ log 𝑙𝑖𝑚𝑦 = −1
𝑥→0
1 1
⇒ 𝑙𝑖𝑚𝑦 = 𝑒 =
𝑥→0 𝑒

Summary
We learnt about the concept of going close to a number from all possible directions. Here we are
dealing with real numbers only so there are only two directions i.e. left and right. The limit of a
function was defined more rigorously using the epsilon delta definition. Since the epsilon delta
definition is tedious to apply, so some properties that can be proved by basic definition come
handy to evaluate the limits of various composite functions. Moreover we learnt the concept of the
limit at infinity and of infinite limits in eight different cases.

Key words
limit, epsilon-delta definition of limit, limit at infinity, infinite limits

Review Questions
1. When do you say that a function 𝑓(𝑥) is approaching to infinity as 𝑥 → ∞?
2. The L’Hopital rule is given as

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Notes
𝑓(𝑥) 𝑓′ (𝑥)
𝑙𝑖𝑚 = 𝑙𝑖𝑚
(a) 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔′ (𝑥)

𝑓(𝑥) 𝑓(𝑎)
(b) 𝑙𝑖𝑚 =
𝑥→𝑎 𝑔(𝑥) 𝑔(𝑎)

𝑓(𝑥) 𝑓′ (𝑎)
(c) 𝑙𝑖𝑚 =
𝑥→𝑎 𝑔(𝑥) 𝑔′ (𝑎)

𝑓(𝑥) 𝑓(𝑥)
(d) 𝑙𝑖𝑚 = 𝑙𝑖𝑚[ ]
𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔(𝑥)

tan 5𝑥
3. Evaluate 𝑙𝑖𝑚𝜋
𝑥→ 2 tan 𝑥
sin 𝑥
4. 𝑙𝑖𝑚 is
𝑥→0 𝑥
(a) 0 (b) 1 (c) -1 (d) undefined
1 𝑛
5. 𝑙𝑖𝑚 (1 + ) is given as
𝑛→∞ 𝑛
(a) 1 (b) 2 (c) 3 (d) e
1
6. Show that 𝑙𝑖𝑚 =∞
𝑥→0 𝑥 2

Further/Suggested Readings
George B. Thomas Jr., Joel Hass, Christopher Heil & Maurice D. Weir (2018). Thomas’ Calculus
(14th edition). Pearson Education.
Howard Anton, I. Bivens & Stephan Davis (2016). Calculus (10th edition). Wiley India.

https://www.mathsisfun.com/calculus/index.html
https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_
Transcendentals_(Stewart)/04%3A_Applications_of_Differentiation/4.04%3A_Indeterminate_
Forms_and_l%27Hospital%27s_Rule

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Dr. Pratiksha, Lovely Professional University Unit 05: Continuity of a real valued function
Notes

Unit 05: Continuity of a real valued function

CONTENTS
Objectives
Introduction
5.1 Concept of continuity
5.2 Properties of continuous functions
5.3 Intermediate value theorem
5.4 Geometric interpretation of continuity
5.5 Types of discontinuity
5.6 Uniform continuity
Summary
Key Words
Self Assessment
Answers for Self Assessment
Review Questions
Further Readings

Objectives
After studying this unit Students will

 learn about the concept of continuity and its geometric interpretation.

 explore some properties of the continuous functions
 learn various types of discontinuity
 be able to differentiate in continuity and uniform continuity

Introduction
With the word ‘continuity’, the first thing that comes to our mind is the ability to draw a graph
without lifting the pen.In this unit we will learn when can we tag a function as a continuous
function and if a function is continuous then what more can we know about the function. You will
learn about the various types of discontinuities and the concept of uniform continuity also.

5.1 Concept of continuity

Consider a function 𝑓 𝑥 .As the independent variable 𝑥 will change, somehow 𝑓 𝑥 will also
change. The idea of continuity is that if a small change is happening in 𝑥 then a small change must
happen in 𝑓 𝑥 i.e. the change in 𝑓(𝑥) should not be sudden for a small change occurring in 𝑥. Now
here the word ‘small’ is not defined in a complete sense. My idea of small can differ from your idea
of small. Thus a more precise epsilon delta definition is there to address this issue.

Continuity of a function at a point in an interval

A function 𝑓 is said to be continuous at a point 𝑐, if to any 𝜖 > 0 , there corresponds a number 𝛿 > 0
such that
𝑓 𝑐+𝑕 −𝑓 𝑐 <𝜖
for all values of 𝑕, such that 𝑕 < 𝛿.
In a different manner, a function 𝑓(𝑥) is said to be continuous at 𝑐, if ∃ an interval 𝑐 − 𝛿, 𝑐 +
𝛿 around 𝑐, such that for all 𝑥 ∈ (𝑐 − 𝛿, 𝑐 + 𝛿), we have
𝑓 𝑐 −𝜖 <𝑓 𝑥 <𝑓 𝑐 +𝜖

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Notes
A function 𝑓 𝑥 is continuous at an interior point 𝑐 if and only if limx→c f x = 𝑓(𝑐)

Show that 𝑓 𝑥 = 3𝑥 + 1 is continuous at 𝑥 = 1.

Here 𝑓 1 = 4
𝑓 𝑥 − 𝑓 1 = 3𝑥 − 3 = 3 𝑥 − 1
Let 𝜖 > 0 be given, we will show that 𝑓 𝑥 − 𝑓 1 < 𝜖 for some 𝛿 > 0 such that 𝑥 − 1 < 𝛿.
Let 𝑥 > 1
It implies 3 𝑥 − 1 > 0
Now 𝑓 𝑥 − 𝑓 1 = 3 𝑥−1 < 𝜖
𝜖
i.e. if 𝑥 − 1 <
3
𝜖
i.e. if 𝑥 < 1 + ------ (1)
3

Now let 𝑥 < 1

It implies 3 𝑥 − 1 < 0
Now 𝑓 𝑥 − 𝑓 1 = 3 𝑥−1 < 𝜖
𝜖
i.e. if 1 − 𝑥 <
3
𝜖
i.e. if 𝑥 > 1 − -----(2)
3

From (1) and (2) we can say that

𝜖 𝜖
𝑓 𝑥 −𝑓 1 < 𝜖if1 − < 𝑥 < 1 +
3 3
𝜖 𝜖
i.e. if – < 𝑥 − 1 <
3 3

In other words we have found a 𝛿 such that −𝛿 < 𝑥 − 1 < 𝛿 or 𝑥 − 1 < 𝛿

Therefore 𝑓 is continuous at 1.

Prove that 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 is continuous at any point 𝑐 in the domain.

Let 𝜖 > 0 be any number. Consider
𝑓 𝑥 −𝑓 𝑐 = | sin 𝑥 − sin 𝑐|
𝑥+𝑐 𝑥−𝑐
= |2 𝑐𝑜𝑠 𝑠𝑖𝑛 |
2 2
𝑥+𝑐 x−c
= 2 cos 𝑠𝑖𝑛 ----- (1)
2 2
𝑥−𝑐 𝑥−𝑐 𝑥+𝑐
Now sin ≤ and | cos | ≤ 1 ∀ 𝑥and 𝑐.
2 2 2

Now (1) can be written as

𝑥−𝑐
| sin 𝑥 − sin 𝑐| ≤ 2 = 𝑥−𝑐
2
Therefore | sin 𝑥 − sin 𝑐| < 𝜖 when 𝑥 − 𝑐 < 𝜖 = 𝛿
Thus there exists an interval around 𝑐, such that ∀ 𝑥 ∈ 𝑐 − 𝛿, 𝑐 + 𝛿 ,
| sin 𝑥 − sin 𝑐| < 𝜖
Therefore 𝑓 𝑥 = sin 𝑥 is continuous at 𝑐.

𝑥 2 −1
Examine lim𝑥→1
𝑥−1

When 𝑥 ≠ 1 the given function can be written as 𝑦 = 𝑥 + 1.

Let 𝜖 > 0 be any number, however small.
Let 𝑥 > 1 then 𝑦 > 2

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 Unit 05: Continuity of a real valued function
Notes
𝑦 − 2 = 𝑦 − 2 = 𝑥 + 1 − 2 = 𝑥 − 1 < 𝜖if𝑥 < 1 + 𝜖
Therefore there exists an interval (1, 1 + 𝜖) such that 𝑦 − 2 < 𝜖
∴ lim 𝑦 = 2as𝑥 tends to 1+.
Let 𝑥 < 1 then 𝑦 < 2
𝑦 − 2 = 2 − 𝑦 = 2 − 𝑥 − 1 = 1 − 𝑥 < 𝜖if𝑥 > 1 − 𝜖
Therefore there exists an interval (1 − 𝜖, 1) such that 𝑦 − 2 < 𝜖
∴ lim 𝑦 = 2as𝑥 tends to 1−.
Thus from both directions 𝑦 − 2 < 𝜖 whenever 𝑥 − 1 < 𝜖
∴ lim 𝑦 = 2 as 𝑥 → 1.

5.2 Properties of continuous functions

A function is said to be continuous if it is continuous at every point of its domain. Check the
domain of the function and apply the definition of continuity at the suspicious point.
Let us check the function 𝑓 𝑥 = sin2 𝑥 for continuity.
Cleary the set of real numbers is the domain of the function and to check if the function is
continuous or not, we need to check that the function should be continuous on each point of its
domain. Let us consider an arbitrary real number 𝑐 and any 𝜖 > 0. Then
𝑓 𝑥 −𝑓 𝑐 = | sin2 𝑥 − sin2 𝑐| = |sin⁡
(𝑥 + 𝑐)||sin⁡
(𝑥 − 𝑐)|
≤ |sin⁡
(𝑥 − 𝑐)| ≤ |𝑥 − 𝑐|
If 𝑥 − 𝑐 < 𝜖 = 𝛿 then 𝑓 𝑥 − 𝑓 𝑐 <𝜖
Therefore by definition sin2 𝑥 is continuous for 𝑥 = 𝑐∀ 𝑥 as 𝑐 is any number.
Let us now see a piecewise function for its continuity. Let the function be
1
𝑥 when 0 ≤ 𝑥 <
2
1
𝑓 𝑥 = 1 when 𝑥 =
2
1
1−𝑥 when < 𝑥 < 1
2
Here the function is a polynomial function or a constant function in the piecewise domains.
1
The problem of discontinuity can occur at . So let us work out on the left hand limit, the right hand
2
1
limit and the value of the function at .
2
1
The left hand limit is lim𝑥→1− 𝑥 =
2 2

1
The right hand limit is lim 1+ 𝑥=
𝑥→ 2
2

1 1
The value of the function at , 𝑓 =1
2 2

The limit of the function is existing but is not equal to the value of the function, so the function is
1 1
not continuous at 𝑥 = . The point 𝑥 = is very much in the domain of the function, so we can say
2 2
that the function is not continuous.

Theorems on continuous functions

Suppose 𝑓and 𝑔are two functions defined in a neighbourhood of the point 𝑎. Then, if lim 𝑓(𝑥)and
𝑥→𝑎

lim 𝑔(𝑥)are well-defined, we have the following:

𝑥→𝑎

(1) lim⁡
(𝑓 𝑥 + 𝑔(𝑥))is defined, and equals the sum of the values lim 𝑓 𝑥 and lim 𝑔(𝑥).
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

(2) lim (𝑓 𝑥 − 𝑔 𝑥 )is defined, and equals lim 𝑓(𝑥)− lim 𝑔 𝑥 .

𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

(3) lim 𝑓(𝑥)𝑔(𝑥) is defined, and equals the product lim 𝑓 𝑥 lim 𝑔(𝑥).
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

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Notes
The scalar multiples result basically states that if lim 𝑓 𝑥 exists, and for any real number 𝛼
𝑥→𝑎

lim 𝛼𝑓 𝑥 = 𝛼lim 𝑓(𝑥)

𝑥→𝑎 𝑥→𝑎

(4) lim 𝑓 𝑥 andlim 𝑔(𝑥) exist and if lim 𝑔 𝑥 ≠ 0, then

𝑥→𝑎 𝑥→𝑎 𝑥 →𝑎

𝑓 𝑥 lim 𝑓 𝑥
lim = 𝑥→𝑎
𝑥→𝑎 𝑔 𝑥 lim 𝑔 𝑥
𝑥→𝑎

The notions of sum, difference, product and quotient of functions can be rewritten as:
lim⁡
(𝑓 + 𝑔) 𝑥 = lim⁡𝑓 𝑥 + lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim⁡
(𝑓 − 𝑔) 𝑥 = lim 𝑓 𝑥 − lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim⁡
(𝑓. 𝑔) 𝑥 = lim⁡𝑓 𝑥 . lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim (𝑓/ 𝑔 ) 𝑥 = lim 𝑓 𝑥 /lim⁡𝑔 𝑥 provided𝑔 𝑎 ≠ 0

𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

You are much familiar with some elementary functions such as the constant function, identity
function, rational function, trigonometric functions, inverse trigonometric functions , exponential
functions and the logarithmic functions.

The domain of continuity of a function is same as the domain of the definition of function.
All the above mentioned elementary functions are continuous in their domain.
For the composed functions, we need to check the continuity every time. Some properties of
continuous functions can be stated as follows.

1. If 𝑓 𝑥 is continuous at 𝑐 and 𝑓 𝑐 ≠ 0 the there exists an open interval 𝑐 − 𝛿, 𝑐 + 𝛿

around 𝑐 such that 𝑓 𝑥 has the sign of 𝑓 𝑐 for every 𝑥 in this interval.
2. If 𝑓 is continuous in a closed interval [𝑎, 𝑏] and 𝑓 𝑎 , 𝑓(𝑏) are of opposite signs, then 𝑓(𝑥)
is zero for atleast one 𝑥 ∈ 𝑎, 𝑏 .
3. If 𝑓 is continuous in [𝑎, 𝑏], then there exist points 𝑐 and 𝑑 in the interval [𝑎, 𝑏] where 𝑓
assumes its greatest and least values 𝑀 and 𝑚 that is,
𝑓 𝑐 = 𝑀and𝑓 𝑑 = 𝑚.

5.3 Intermediate value theorem

This theorem applies to the continuous functions. Using this theorem you can prove the solvability
of the algebraic and transcendental equations.
For example, sin 𝑥 + 𝑥 5 = 0 is a transcendental equation and we can use the intermediate theorem
to know whether it is solvable or not.

INTERMEDIATE VALUE THEOREM: Let 𝑓 be a continuous function on the closed interval

[𝑎, 𝑏]. Assume that 𝑚 is a number (y-value) between 𝑓 𝑎 and𝑓(𝑏). Then there is at least one number
𝑐(x- value) in the interval 𝑎, 𝑏 which satisfies 𝑓 𝑐 = 𝑚.
Assume that a function 𝑓 is a continuous and𝑚 = 0. Then the conditions 𝑓 𝑎 < 0 and𝑓 𝑏 > 0
would lead to the conclusionthat the equation 𝑓 𝑥 = 0 is solvable for 𝑥, i.e., 𝑓 𝑐 = 0.
Intermediate Value Theorem guarantees the existence of a solution, but not what the solution is.
Steps to solve a problem:
1. Define a function 𝑦 = 𝑓 𝑥 .
2. Establish that 𝑓 is continuous.
3. Choose an interval 𝑎, 𝑏 .
4. Define a number (y-value) 𝑚.
5. Establish that there exists a value 𝑐, in 𝑎, 𝑏 such that 𝑓 𝑐 = 𝑚.

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 Unit 05: Continuity of a real valued function
Notes
Now let us use the intermediate value theorem to prove that the equation is solvable on the given
interval in the following examples.

3𝑥 5 − 4𝑥 2 = 3 on[0,2]
Let 𝑓 𝑥 = 3𝑥 5 − 4𝑥 2 − 3
𝑓(𝑥)is continuous for every 𝑥 as it is a polynomial function.
Now here 𝑓 0 = −3, 𝑓 2 = 77
Let 𝑚 = 3 as 𝑓 0 < 𝑚 < 𝑓(2)
By intermediate value theorem, we can conclude that there exists 𝑐 ∈ [0,2] such that 𝑓 𝑐 = 𝑚
i.e.3𝑐 5 − 4𝑐 2 + 3 = 3
or𝑐 2 3𝑐 3 − 4 = 0
4 1/3
or𝑐 = 0,0,
3

Here all the values of 𝑐 are lying in the given interval. In fact if only one value lies in the interval,
that itself is sufficient to say that the equation is solvable. If we draw the function on 𝑥𝑦 −plane, the
results are quite obvious.

𝑥 2 − 4𝑥 3 + 1 = 𝑥 − 7
Let 𝑓 𝑥 = 𝑥 2 − 4𝑥 3 + 1 − 𝑥 + 7 = −4𝑥 3 + 𝑥 2 − 𝑥 + 8
Here 𝑓 𝑥 is continuous for all 𝑥.
Now in this problem the interval is not given as in the previous example. Therefore by hit and trial
we can look for two values 𝑥 such that the value of the function at those values are of opposite
signs.

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 Calculus
Notes
𝑓 0 =8
𝑓 2 = −22
Let 𝑚 = 0 (any number between -22 and 8 can be chosen as 𝑚)
Clearly all the assumptions of intermediate value theorem are met. Therefor there exists a 𝑐 ∈ 0,2
such that 𝑓 𝑐 = 𝑚
i.e. −4𝑐 3 + 𝑐 2 − 𝑐 + 8 = 0 and this equation is solvable.
If we check it by actually drawing the graph, we can easily see that the intermediate value theorem
is getting satisfied in the said interval.

𝑥 3 + 2 = 𝑠𝑖𝑛𝑥
Now this equation is a transcendental one. and no interval is given.
Let 𝑓 𝑥 = 𝑥 3 + 2 − sin 𝑥
The function 𝑓 𝑥 is the sum of continuous functions so it is a continuous function for all 𝑥.
𝑓 0 =2
𝑓 −𝜋 = −29
Choosing 𝑚 such that −29 < 𝑚 < 2.
Let 𝑚 = 0
∴ ∃ 𝑐 ∈ −𝜋, 0 such that 𝑓 𝑐 = 𝑚
i.e.𝑐 3 + 2 − sin 𝑐 = 0
Therefore the equation is solvable. And we can verify this by actually plotting the graph which is as
follows:

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 Unit 05: Continuity of a real valued function
Notes

5.4 Geometric interpretation of continuity

Continuity at a point𝑐 can be defined for a function𝑓 on an open interval containing𝑐. We may say
that 𝑓is continuous at 𝑐 if 𝑓 𝑥 tends to 𝑓 𝑐 as 𝑥 tends to 𝑐. Or in plane words, the function 𝑓 is
continuous if the difference in 𝑥 and 𝑐 is small, the difference in 𝑓 𝑥 and 𝑓(𝑐) will also be small.
That cannot be abrupt.
In simple words,
(i) Function 𝑓will be continuous at 𝑥 = 𝑐if there is no break in the graph of the function at the point
(𝑐, 𝑓(𝑐)).
(ii) In an interval, function is said to be continuous if there is no break in the graph of the function
in the entire interval.

5.5 Types of discontinuity

Discontinuity of a function
A function is said to be discontinuous at a point of its domain if it is not continuous at that point.
Moreover, that point is called the point of discontinuity of the function.
Two possibilities:

1. The limit of the functions exists as x tends to c but is different from the value of the
function at c.
2. The limit of the function does not exist at c.
On this basis, we can classify the discontinuities as follows:

1. Removable discontinuity
2. Jump discontinuity (Discontinuity of the first kind)
3. Discontinuity of the first kind from the left
4. Discontinuity of the first kind from the right
5. Discontinuity of the second kind ( Non removable or essential discontinuity)
6. Discontinuity of the second kind from the left

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 Calculus
Notes
7. Discontinuity of the second kind from the right

Removable discontinuity
If lim𝑥→𝑐 𝑓 𝑥 exists but is not equal to 𝑓(𝑐) (which may or may not exist), then that discontinuity is
called removable because we can redefine the function so that the function becomes continuous at
point 𝑐.
i.e.lim𝑥 →𝑐 − 𝑓 𝑥 = lim𝑥→𝑐 + 𝑓 𝑥 ≠ 𝑓(𝑐)
sin 𝑥
if 𝑥 ≠ 0
For instance consider the function 𝑓 𝑥 = 𝑥
0 if 𝑥 = 0
Clearly the limit of the function exists and is 1. But the value of the function at 𝑥 = 0 is not 1.
Thus the function has a removable discontinuity at 𝑥 = 0.
It means that we can redefine the function to remove this discontinuity by writing
sin 𝑥
𝑓 𝑥 = if 𝑥 ≠ 0
𝑥
1 if 𝑥 = 0
Jump discontinuity
This discontinuity is also called discontinuity of first kind. If lim𝑥→𝑐 − 𝑓 𝑥 and lim𝑥→𝑐 + 𝑓 𝑥 both
exist but are not equal, then we get a jump discontinuity.
i.e.lim𝑥 →𝑐 − 𝑓 𝑥 ≠ lim𝑥→𝑐 + 𝑓 𝑥
1 if 𝑥 > 0
For instance 𝑓 𝑥 =
−1 if 𝑥 < 0
Clearly the left hand limit is -1 and right hand limit is +1 and you can see a jump right there near
zero, hence the name ‘jump discontinuity’.

Discontinuity of the first kind from the left

If lim𝑥→𝑐 − 𝑓 𝑥 ≠ 𝑓 𝑐 = lim+ 𝑓 𝑥
𝑥→𝑐

Discontinuity of the first kind from the right

If lim𝑥→𝑐 − 𝑓 𝑥 = 𝑓 𝑐 ≠ lim+ 𝑓 𝑥
𝑥→𝑐

Discontinuity of the second kind (Non removable or essential discontinuity)

If neither lim𝑥→𝑐 − 𝑓 𝑥 nor lim𝑥→𝑐 + 𝑓 𝑥 exists.

Discontinuity of the second kind from the left

If lim𝑥→𝑐 − 𝑓 𝑥 does not exist.

Discontinuity of the second kind from the right

If lim𝑥→𝑐 + 𝑓 𝑥 does not exist.

𝑥
Consider a function 𝑓 𝑥 = 51−𝑥 2
Clearly the function is not defined at 1 and -1. But we can see how the function will behave as 𝑥
approaches to 1 from both possible directions.
𝑥
The left hand limit = lim𝑥→1− 51−𝑥 2
Put 𝑥 = 1 − 𝑕. 𝑕 → 0as𝑥 → 1−
So the function can now be written as
𝑥 1−𝑕
lim− 51−𝑥 2 = lim 51−(1−𝑕 )2
𝑥→1 𝑕→0

1−𝑕
1−h 1
= lim 5𝑕 (2−𝑕 ) (Here tends to as 𝑕 tends to zero.)
𝑕→0 2−h 2
1
= lim 52𝑕
𝑕→0

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 Unit 05: Continuity of a real valued function
Notes
=∞
𝑥
The right hand limit is lim𝑥→1+ 51−𝑥 2 and
𝑥 1+𝑕
lim+ 51−𝑥 2 = lim 51−(1+𝑕 )2
𝑥→1 𝑕→0
1+𝑕
= lim 5 −2𝑕 −𝑕 2
𝑕→0
1
= lim 5− 2𝑕 = 0
𝑕→0

Similarly at the point 𝑥 = −1, the left hand limit is

𝑥 −1−𝑕
lim − 51−𝑥 2 = lim 51−(−1−𝑕 )2 = ∞
𝑥→−1 𝑕→0

And the right hand limit is

𝑥 −1+𝑕
lim 51−𝑥 2 = lim 51− −1+𝑕 2 = 0
𝑥→ −1+ 𝑕→0

Therefore in this example the point -1 is a discontinuity of the second kind from the left and 1 is
also a discontinuity of the second kind from the left.

5.6 Uniform continuity

To understand uniform continuity let us recapitulate that a function 𝑓is said to be continuous on
𝑹if for every 𝑐 ∈ 𝑹and for every 𝜖 > 0, there exists a 𝛿 > 0 such that for every 𝑥 ∈ 𝑹with 𝑥 − 𝑐 < 𝛿
we have 𝑓 𝑥 − 𝑓 𝑐 < 𝜖. Here 𝛿 can depend upon 𝜖 and 𝑐.
On the other hand a function 𝑓 is said to be uniformly continuous on 𝑹if for every 𝜖 > 0, there
exists a 𝛿 > 0 such that for every 𝑥, 𝑦 ∈ 𝑹with 𝑥 − 𝑦 < 𝛿 we have 𝑓 𝑥 − 𝑓 𝑦 < 𝜖. Here 𝛿 can
depend upon 𝜖.
The uniform continuity is a global concept. Here you have one single rectangle for the whole
domain whereas the continuity of a function at a point is a local concept and the size of the
rectangle will highly depend on the value of 𝑐.
We will see two theorems without proof to have a better idea of uniform continuity.
Theorem 1: Every uniformly continuous function on an interval is continuous on that interval but
the converse is not true.
Theorem 2: If a function is continuous on a closed interval, then it is uniformly continuous on that
closed interval.

𝑥
Is the function 𝑓 𝑥 = uniformly continuous for 𝑥 ∈ [0,2]?
𝑥+1

Let 𝑥, 𝑦 be two arbitrary points in 0,2 .Then𝑥 ≥ 0, 𝑦 ≥ 0

or𝑥 + 1 ≥ 1, 𝑦 + 1 ≥ 1
or 𝑥 + 1 𝑦 + 1 ≥ 1
1
or ≤1
𝑥+1 𝑦 +1

𝑥 𝑦 𝑥−𝑦
Now 𝑓 𝑥 − 𝑓 𝑦 = − = ≤ |𝑥 − 𝑦|
𝑥+1 𝑦 +1 𝑥+1 𝑦 +1

Let 𝜖 > 0 be given. Choosing 𝛿 = 𝜖 we get

𝑓 𝑥 −𝑓 𝑦 < 𝜖when ever 𝑥 − 𝑦 < 𝛿 for every 𝑥, 𝑦 ∈ [0,2]
The same problem can be solved by using the second theorem also. The only problem point for the
function is {−1} and that is not in the domain. So all conditions are getting fulfilled of Theorem 2
and that implies that the given function is uniformly continuous in 0,2 .

1
Show that the function 𝑓 𝑥 = 2 is uniformly continuous on [𝑎, ∞) where 𝑎 > 0, but not
𝑥
uniformly continuous on 0, ∞ .
Let 𝑥, 𝑦 ≥ 𝑎 > 0 be two arbitrary numbers in 0, ∞ .

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 Calculus
Notes
1 1
𝑓 𝑥 −𝑓 𝑦 = 2 − 2
𝑥 𝑦
1 1 1 1
= − +
𝑥 𝑦 𝑥 𝑦
2 𝑦−𝑥
≤
𝑎 𝑥𝑦
2
≤ 𝑥−𝑦
𝑎3
𝜖𝑎 3
Let 𝜖 > 0 be given. Choose 𝛿 = . Then,
2

𝑓 𝑥 −𝑓 𝑦 < 𝜖when 𝑥 − 𝑦 < 𝛿 for all 𝑥, 𝑦 ≥ 𝑎.

Therefore 𝑓 is uniformly continuous on 𝑎, ∞ .
To show that 𝑓 is not uniformly continuous on [0, ∞) let us take two numbers in the interval [0, ∞)
as follows:
1 1
𝑥1 = and𝑥2 = be two numbers.
𝑛 𝑛+1
1 1
Now 𝑓 𝑥1 − 𝑓 𝑥2 = −
𝑥 12 𝑥 22

= 𝑛− 𝑛+1 =1
1 1
And 𝑥1 − 𝑥2 = −
𝑛 𝑛+1

1
=
𝑛 𝑛+1 𝑛+1+ 𝑛
1 1
< = = 𝛿 (say)
𝑛 .2 𝑛 2𝑛
1 1 1
Let 𝜖 = and 𝛿 be any positive number such that 𝑛 > or < 𝛿.
2 2𝛿 2𝑛

Therefore 𝑓 𝑥1 − 𝑓 𝑥2 > 𝜖 when 𝑥1 − 𝑥2 < 𝛿

∴ 𝑓is not uniformly continuous on 0, ∞ .

Summary
In this unit we have learnt about the technical definition of a continuous function and its various
properties.

 A function 𝑓(𝑥) is said to be continuous at 𝑐, if ∃ an interval 𝑐 − 𝛿, 𝑐 + 𝛿 around 𝑐, such

that for all 𝑥 ∈ (𝑐 − 𝛿, 𝑐 + 𝛿), we have
𝑓 𝑐 −𝜖 <𝑓 𝑥 <𝑓 𝑐 +𝜖

 For two continuous functions, the following properties hold.

lim⁡
(𝑓 + 𝑔) 𝑥 = lim⁡𝑓 𝑥 + lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim⁡
(𝑓 − 𝑔) 𝑥 = lim 𝑓 𝑥 − lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim⁡
(𝑓. 𝑔) 𝑥 = lim⁡𝑓 𝑥 . lim⁡𝑔 𝑥
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

lim (𝑓/ 𝑔 ) 𝑥 = lim 𝑓 𝑥 /lim⁡𝑔 𝑥 provided𝑔 𝑎 ≠ 0

𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

 A function is said to be discontinuous at a point of its domain if it is not continuous at that

point. Moreover, that point is called the point of discontinuity of the function.
 We can classify the discontinuities mainly as:
1. Removable discontinuity
2. Jump discontinuity
3. Non removable or essential discontinuity

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 Unit 05: Continuity of a real valued function
Notes
 The intermediate value theorem states that if a continuous function attains two values, it
must also attain all values in between these two values.
 A function 𝑓 is said to be uniformly continuous on 𝑹if for every 𝜖 > 0, there exists a 𝛿 > 0
such that for every 𝑥, 𝑦 ∈ 𝑹with 𝑥 − 𝑦 < 𝛿 we have 𝑓 𝑥 − 𝑓 𝑦 < 𝜖. Here 𝛿 can depend
upon 𝜖.
 Every uniformly continuous function on an interval is continuous on that interval but the
converse is not true.
 If a function is continuous on a closed interval, then it is uniformly continuous on that
closed interval.

Key Words
continuity, discontinuity, intermediate value theorem, uniform continuity

Self Assessment
1. Which of the following is a continuous function?
(a) Constant function
(b) Polynomial function
(c) Sine function
(d) All of the above
2. To verify that any equation is solvable or not, which theorem must be used?
(a) Mean value theorem
(b) Rolle’s theorem
(c) Intermediate value theorem
(d) None of these
3. If a function is continuous, it is definitely uniformly continuous.(True/False)
4. If you can redefine a function so that it becomes continuous, what kind of discontinuity
are you tackling?
(a) Jump
(b) Removable
(c) Discontinuity of first kind
(d) Discontinuity of second kind
𝑥−2
5. lim𝑥→2 is
𝑥 2 −4

(a) 1
(b) 4
1
(c)
4
(d) None of these
6. Which of these functions is not uniformly continuous on (0,1)?
(a) 𝑥 2
𝟏
(b)
𝒙𝟐
(c) sin 𝑥
𝑠𝑖𝑛𝑥
(d)
𝑥
𝑥− 𝑎
7. Find lim𝑥→𝑎 for 𝑏 > 0.
𝑥−𝑎
(a) 0
(b) ∞
(c) 𝑎
1
(d)
2 𝑎
8. Which of the following is not a continuous function?
(a) [𝒙]

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 Calculus
Notes
(b) |𝑥|
(c) 𝑥 2
1
(d) ,𝑥 ≠ 0
𝑥
𝑥 2 −4
9. If a function 𝑓 is continuous for all real numbers and if 𝑓 𝑥 = when 𝑥 ≠ −2, then
𝑥+2
𝑓(−2) is equal to
(a) -4
(b) -2
(c) -1
(d) 0
10. Which of the following functions are continuous at 𝑥 = 1
I. ln 𝑥
II. 𝑒𝑥
III. ln 𝑒 𝑥 − 1
(a) I only
(b) I and II only
(c) II and III only
(d) I, II and III
1 1
−𝑥
𝑒 𝑥 +𝑒
11. The function 𝑓 𝑥 = 1 1 , 𝑥 ≠ 0, 𝑓 0 = 1 has 𝑥 = 0 as a
𝑒 𝑥 −𝑒 −𝑥
(a) removable discontinuity
(b) jump discontinuity
(c) discontinuity of second kind from left
(d) discontinuity of second kind from right
12. Discontinuity of second kind happens when
(a) the left hand limit does not exist
(b) the right hand limit does not exist
(c) both the left hand and right hand limits do not exist
(d) neither the left hand nor right hand limits exist
13. Discontinuity of the second kind is also known as
(a) removable discontinuity
(b) essential discontinuity
(c) jump discontinuity
(d) non-essential discontinuity
14. Jump discontinuity is a
(a) discontinuity of the first kind
(b) discontinuity of the first kind from left
(c) discontinuity of the first kind from right
(d) discontinuity of the second kind
15. Which of the following is/are true?
I. Every uniformly continuous function on an interval is continuous on that interval
and conversely.
II. If a function is continuous on an open interval, then it is uniformly continuous on
that closed interval.
(a) Only I is true
(b) Only II is true
(c) Both I and II are true
(d) None is true

Answers for Self Assessment

1. D 2. C 3. False 4. B 5. C

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 Unit 05: Continuity of a real valued function
Notes
6. B 7. D 8. A 9. A 10. D

11. B 12. D 13. B 14. A 15. D

Review Questions
1. Prove that a constant function is a continuous function.
2. State the intermediate value theorem with an example.
3. Show that the function 𝑓 𝑥 = 6𝑥 − 5 is continuous at 𝑥 = 0.
4. Discuss the continuity of the following function:
3𝑥 − 5, if 𝑥 ≠ 1
𝑓 𝑥 =
2, if 𝑥 = 1
5. Determine the values of 𝐴 and 𝐵 so that the following function is continuous for all values
of x.
𝐴𝑥 − 𝐵, if 𝑥 ≤ −1
𝑓 𝑥 =
2𝑥 2 + 3𝐴𝑥 + 𝐵, if -1< 𝑥 ≤ 1

and𝑓 𝑥 = 4, if 𝑥 > 1
6. Verify if the equation 𝑥 3 = cos 𝑥 − 2 is solvable or not?
𝑥−5
, if 𝑥 ≠ 5
7. Examine the continuity of 𝑓 𝑥 = 𝑥−5 and discuss in case of any discontinuity.
1, if 𝑥 = 5
𝑥
8. Check if the function is uniformly continuous on 0,2 .
𝑥+2
𝑥−5
, if 𝑥 > 5
9. Examine the continuity of 𝑓 𝑥 = 𝑥−5
1, if 𝑥 ≤ 5
10. Discuss the kind ofdiscontinuity for the following function:
3𝑥 − 5, if 𝑥 ≠ 1
𝑓 𝑥 =
2, if 𝑥 = 1

Further Readings
George B. Thomas Jr., Joel Hass, Christopher Heil& Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens& Stephan Davis (2016).Calculus (10th edition).Wiley India.

Web Links
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins
_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits_Continuity_and_Differenti
ability

50 LOVELY PROFESSIONAL UNIVERSITY

 Unit 06: Differentiability of a Real Valued Function Notes
Dr. Pratiksha, Lovely Professional University

Unit 06: Differentiability of a Real Valued Function

CONTENTS
Objectives
Introduction
6.1 Derivability and derivative
6.2 Geometrical interpretation of differentiability
6.3 Relation between differentiability and continuity
6.4 Differentiability and monotonicity
Summary
Key Words
Self Assessment
Answers for Self Assessment
Review Questions
Further Readings

Objectives
Students will

• learn about the concept of differentiability

• explore the geometric interpretation of differentiability
• understand the relation in differentiability and continuity
• analyze the connection of differentiability and monotonicity

Introduction
After understanding the concept of limit, we can move on to know what is meant by
differentiability of a real valued function. As we know, calculus is all about the study of the change.
The general idea that comes to mind is the average rate of change. We want to say how fast we are,
and to know that, we go for finding the average speed, so if you want to compare two persons, you
just see their average speed or average velocity. Similarly, if you want to see where the bend is
sharper, you would like to see the curvature. To explain the phenomena in more detail, we would
like to know the velocity ‘at a particular point’ or the curvature ‘at a particular point’. Here we are
basically interested in the local change or the instantaneous rate of change. The average rate of
change is kind of global phenomenon, we are telling something for the whole period of time or for
the whole domain in general. In the first case, we need to understand something called
differentiability which is an instantaneous phenomenon or alocal phenomenon.

6.1 Derivability and derivative

Consider a function 𝑓: 𝐴 → 𝐵. Let 𝑐 ∈ 𝐴 be any point in the domain 𝐴 and 𝐵 is a set of real numbers.
For ℎ > 0, 𝑐 + ℎ and 𝑐 − ℎ lie on the right and left of 𝑐 respectively. The value of the function at 𝑐
and 𝑐 + ℎ are respectively 𝑓(𝑐)and 𝑓(𝑐 + ℎ). Thus we can say that change in 𝑥 is ℎ and change in
𝑓(𝑥) is 𝑓(𝑐 + ℎ) − 𝑓(𝑐).
𝑓(𝑐+ℎ)−𝑓(𝑐)
The average rate of change of the function w.r.t. the independent variable 𝑥 is . As the
ℎ
𝑓(𝑐+ℎ)−𝑓(𝑐)
value of ℎ approaches to zero, the expression may tend to a limit. The limit if exists, is
ℎ
called the derivative of the function 𝑓 at point 𝑐, and is denoted by 𝑓 ′ (𝑐)and
𝑓(𝑐 + ℎ) − 𝑓(𝑐)
𝑓 ′ (𝑐) = lim
ℎ→0 ℎ
If the derivative of the function takes a finite value, the function is called finitely derivable at 𝑐.

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The left hand derivative is given as
𝑓(𝑐 + ℎ) − 𝑓(𝑐)
𝑓 ′ (𝑐 − ) = lim−
ℎ→0 ℎ
and the right hand derivative can be written as
𝑓(𝑐 + ℎ) − 𝑓(𝑐)
𝑓′(c + ) = lim
ℎ→0+ ℎ
The function 𝑓 is derivable if both the left and right hand derivatives exist and are equal.

Prove that the function 𝑓(𝑥) = 𝑥 2 is derivable at 𝑥 = 1.

Here the point of interest is 𝑐 = 1, so we check the functional value at 1 and 1 + ℎ.
𝑓(1) = 1
𝑓(1 + ℎ) = 1 + 2ℎ + ℎ2
𝑓(𝑐+ℎ)−𝑓(𝑐)
To check the derivability, we need to check if the limit lim exists or not!
ℎ→0 ℎ

𝑓(1+ℎ)−𝑓(1) 2ℎ+ℎ2
Now lim = lim = lim 2 + ℎ = 2
ℎ→0 ℎ ℎ→0 ℎ ℎ→0

Thus the limit exists and hence the function is derivable at 𝑥 = 1.

Check the differentiability of the modulus function at the zero.

Let the function be 𝑓(𝑥) = |𝑥|.
The domain of the function is the set of all real numbers. Here the point of interest is 𝑐 = 0. So we
will check if the limit
𝑓(0+ℎ)−𝑓(0)
lim exists or not.
ℎ→0 ℎ

𝑓(0 + ℎ) − 𝑓(0) |ℎ|

lim = lim
ℎ→0 ℎ ℎ→0 ℎ
Since the modulus function is appearing in the function, whose limit is to be calculated, we need to
apply the definition of modulus function which is as follows:
ℎ, ℎ>0
|ℎ| = {
−ℎ, ℎ<0
The left hand derivative is given as
𝑓(0 + ℎ) − 𝑓(0) |ℎ|
𝑓 ′ (0− ) = lim− = lim = −1
ℎ→0 ℎ ℎ→0 ℎ

And the right hand derivative is given as

𝑓(0 + ℎ) − 𝑓(0) |ℎ|
𝑓 ′ (0+ ) = lim+ = lim = +1
ℎ→0 ℎ ℎ→0 ℎ

Clearly the limit does not exist at 𝑥 = 0. Therefor 𝑓 ′ (0) does not exist or we can say that the
modulus function is not differentiable at 𝑥 = 0.

Find the derivative of 𝑓(𝑥) = √𝑥.

The domain of the function is the set of all non-negative real numbers. So we need to check the
differentiability of the function for all 𝑥 > 0 and at 𝑥 = 0.
Let 𝑥 > 0. We can write

𝑓(𝑥 + ℎ) − 𝑓(𝑥) √𝑥 + ℎ − √𝑥
=
ℎ ℎ
√𝑥+ℎ−√𝑥
(√𝑥 + ℎ + √𝑥)
ℎ
=
√𝑥 + ℎ + √𝑥

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1
=
√𝑥 + ℎ + √𝑥
𝑓(𝑥+ℎ)−𝑓(𝑥) 1
Therefore lim = provided 𝑥 > 0 .
ℎ→0 ℎ 2√𝑥

To check the differentiability at 𝑥 = 0, we can examine that

𝑓(0 + ℎ) − 𝑓(0) √ℎ 1
lim = lim = lim
ℎ→0 ℎ ℎ→0 ℎ ℎ→0 √ℎ

1 1
Clearly, lim+ → ∞ and lim− is not defined.
ℎ→0 √ℎ ℎ→0 √ℎ
1
So we can say that 𝑓 ′ (𝑥) = ∀ 𝑥 ∈ (0, ∞)
2√𝑥

Check if the function 𝑓(𝑥) = 𝑥|𝑥| is derivable at the origin.

Since there is modulus function involved in the definition of the function, we can simplify it first.
𝑥 2, 𝑥≥0
It can be written as 𝑓(𝑥) = { 2
−𝑥 , 𝑥<0
The point of interest is the origin. So let’s find the left hand and right hand derivatives at origin.
𝑓(𝑥) − 𝑓(0) −𝑥 2
𝑓 ′ (0− ) = lim− = lim− = lim− − 𝑥 = 0
𝑥→0 𝑥−0 ℎ→0 𝑥 ℎ→0

𝑓(𝑥) − 𝑓(0) 𝑥2
𝑓 ′ (0+ ) = lim+ = lim+ = lim+𝑥 = 0
𝑥→0 𝑥−0 ℎ→0 𝑥 ℎ→0

Therefore 𝑓 ′ (0− ) = 𝑓 ′ (0+ ) and it implies that 𝑓 is derivable at the origin.

6.2 Geometrical interpretation of differentiability

Recall that if 𝑦 = 𝑓(𝑥), then, for any real number Δ𝑥,
Δ𝑦 𝑓(𝑥 + Δ𝑥) − 𝑓(𝑥)
=
Δ𝑥 Δ𝑥

is the average rate of change of 𝑦 with respect to 𝑥 over the interval [𝑥, 𝑥 + Δ𝑥]. Now if the graph of
Δ𝑦
𝑦 is a straight line, that is, if 𝑦 = 𝑚𝑥 + 𝑏 for some real numbers 𝑚and 𝑏, then = 𝑚, the slope of
Δ𝑥
Δ𝑦
the line. In fact, a straight line is characterized by the fact is the same for any values of 𝑥and Δ𝑥.
Δ𝑥
Δ𝑦
Moreover, remains the same when Δ𝑥 is infinitesimal; that is, the derivative of 𝑦 with respect to 𝑥
Δ𝑥
Δ𝑦
is the slope of the line. For other differentiable functions 𝑓 , the value of depends upon both 𝑥and
Δ𝑥
Δ𝑦
Δ𝑥 . However, for infinitesimal values of Δ𝑥, the shadow of , that is, the derivative d𝑦
Δ𝑥 d𝑥
, depends on
d𝑦
𝑥 alone. Hence it is reasonable to think of as the slope of the curve 𝑦 = 𝑓(𝑥)at a point𝑥. Whereas
d𝑥
the slope of a straight line is constant from point to point, for other differentiable functions the
value of the slope of the curve will vary from point to point. If 𝑓 is differentiable at a point 𝑎, we
call the line with slope 𝑓′(𝑎)passing through (𝑎, 𝑓(𝑎))the tangent line to the graph of 𝑓at (𝑎, 𝑓(𝑎)).

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That is, the tangent line to the graph of 𝑦 = 𝑓(𝑥) at 𝑥 = 𝑎 is the line with equation
𝑦 = 𝑓 ′ (𝑎) (𝑥 − 𝑎) + 𝑓(𝑎)
Hence a tangent line to the graph of a function 𝑓 is a line through a point on the graph of 𝑓whose
slope is equal to the slope of the graph at that point.

Kinematic interpretation of differentiability

The motion of a particle along a straight line can be written as 𝑠 = 𝑓(𝑡), where 𝑠 is the distance of
the particle at point 𝑃 from a fixed point of reference on the line,at time 𝑡. Let after some time Δ𝑡,
Δ𝑠
the body covers a distance 𝑠 + Δ𝑠 and is at point 𝑄 now. Clearly represents the average velocity
Δ𝑡
for this interval of time and it approximates the actual velocity at 𝑃. Clearly as this interval of time
gets smaller and smaller, the approximate value of the velocity gets better and better to the actual
value.
𝑓(𝑡+Δ𝑡)−𝑓(𝑡) Δ𝑠 𝑑𝑠
i.e. lim = lim ( ) = = velocity (v) at time 𝑡
Δ𝑡→0 Δ𝑡 Δ𝑡→0 Δ𝑡 𝑑𝑡

Similarly the instantaneous rate of change of velocity is interpreted as acceleration.

𝑣(𝑡+Δ𝑡)−𝑣(𝑡) Δ𝑣 𝑑𝑣
lim = lim ( ) = = acceleration (a) at time 𝑡
Δ𝑡→0 Δ𝑡 Δ𝑡→0 Δ𝑡 𝑑𝑡

The third and fourth derivatives of distance w.r.t. time are called jerk and jounce.
𝑎(𝑡+Δ𝑡)−𝑎(𝑡) Δ𝑎 𝑑𝑎
lim = lim ( ) = = jerk (j) at time 𝑡
Δ𝑡→0 Δ𝑡 Δ𝑡→0 Δ𝑡 𝑑𝑡

𝑗(𝑡+Δ𝑡)−𝑗(𝑡) Δ𝑗 𝑑𝑗
lim = lim ( ) = = jounce (J) at time 𝑡
Δ𝑡→0 Δ𝑡 Δ𝑡→0 Δ𝑡 𝑑𝑡

You can see and feel all these changes physically in the real world. But that will not be the case with
all the functions other than the distance function.
The derivative of various functions have been developed through the ab-initio definition. Two
examples are given.

Derivative of the function 𝒇(𝒙) = 𝒌 where 𝒌 is a given number:

𝑦=𝑘
𝑦 + Δ𝑦 = 𝑘
Δ𝑦 0
= =0
Δ𝑥 Δ𝑥
d𝑦 Δ𝑦
= lim = 0for every 𝑥 ∈ Set of real numbers
d𝑥 Δ𝑥→0 Δ𝑥

Derivative of the function 𝒇(𝒙) = 𝒙𝒏 where 𝒏 is a natural number:

𝑦 = 𝑥𝑛
𝑦 + Δ𝑦 = (𝑥 + Δ𝑥)𝑛
𝑛(𝑛 − 1) 𝑛−2
= 𝑥 𝑛 + 𝑛𝑥 𝑛−1 Δ𝑥 + 𝑥 (Δ𝑥)2 + ⋯ + (Δ𝑥)𝑛
2

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Δ𝑦 𝑛(𝑛 − 1) 𝑛−2
= 𝑛𝑥 𝑛−1 + 𝑥 Δ𝑥 + ⋯ + (Δ𝑥)𝑛−1
Δ𝑥 2
d𝑦 Δ𝑦
= lim = 𝑛𝑥 𝑛−1
d𝑥 Δ𝑥→0 Δ𝑥
Once the basic understanding is there, we can use the formulae of differentiation directly. You have
studied and used them already still the lists of various formulae are presented below for your
revision.

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Notes Calculus

6.3 Relation between differentiability and continuity

In a layman language, if you want to check the differentiability of a function at a point, just by
looking at its graph, you must zoom the figure at that point (in your mind, if not possible
otherwise!) If you see a straight line at that point, the function is differentiable and otherwise it is
non-differentiable. So corners or pointy edges in a graph imply non differentiable nature of the
function while smooth curve suggests a differentiable function.
Similarly the concept of continuity in a domain relates to the graph without any kink or cut or
break in that domain.
Theorem: If f is finitely derivable at c, then f is also continuous at c.

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𝑓 (𝑐+h)−𝑓(𝑐)
Let 𝑓 be a finitely derivable function at 𝑐, so that the expression
ℎ
→ a finite limit asℎ →
0.
We can write
𝑓(𝑐 + h) − 𝑓(𝑐)
𝑓(𝑐 + h) − 𝑓(𝑐) = ℎ
ℎ
𝑓(𝑐 + h) − 𝑓(𝑐)
lim 𝑓(𝑐 + h) − 𝑓(𝑐) = lim lim ℎ
ℎ→0 ℎ→0 ℎ ℎ→0

= 𝑓 ′ (𝑐) ∗ 0 = 0
∴ lim 𝑓(𝑐 + h) − 𝑓(𝑐) = 0
ℎ→0

lim 𝑓(𝑐 + h) = 𝑓(𝑐)

ℎ→0

Alternatively,
lim 𝑓(𝑥) = 𝑓(𝑐)
𝑥→𝑐

It implies that 𝑓 is continuous at 𝑥 = 𝑐.

The converse of this theorem is not necessarily true. One stark example is the absolute function. The
function 𝑓(𝑥) = |𝑥| is continuous on its domain but is not differentiable at the point 0 which is a
part of the domain.
1 1
𝑥(𝑒 − 𝑥 −𝑒 𝑥 )
Examine 𝑓(𝑥) = { 1 1 𝑥 ≠ 0 for the continuity and differentiability at origin.
𝑒 − 𝑥 +𝑒 𝑥
0 𝑥=0
1 1
−𝑥
𝑥(𝑒 −𝑒 𝑥 )
The right hand limit is lim+ 1 1
𝑥→0 𝑒 − 𝑥 +𝑒 𝑥
2
𝑥 (𝑒 − 𝑥 − 1)
= lim+ 2 =0
𝑥→0
𝑒− 𝑥 + 1
1 1
𝑥(𝑒 − 𝑥 −𝑒 𝑥 )
The left hand limit is lim− 1 1
𝑥→0 𝑒 − 𝑥 +𝑒 𝑥
2
𝑥 (1 − 𝑒 𝑥 )
= lim− 2 =0
𝑥→0
1 + 𝑒𝑥
The value of the function at 0 is also zero.
Therefore the given function is continuous at origin.
1 1
𝑓(𝑥)−𝑓(0) 𝑒 − 𝑥 −𝑒 𝑥
The right hand derivative is 𝑓 ′ (0+ ) = lim+ = lim+ 1 1
𝑥→0 𝑥−0 𝑥→0 𝑒 − 𝑥 +𝑒 𝑥
2
−
𝑒 𝑥 −1
= lim+ 2 = −1
𝑥→0 −
𝑒 𝑥 +1
1 1
−𝑥
𝑓(𝑥)−𝑓(0) 𝑒 −𝑒 𝑥
The left hand derivative is 𝑓 ′ (0− ) = lim− = lim− 1 1
𝑥→0 𝑥−0 𝑥→0 𝑒 − 𝑥 +𝑒 𝑥
2
1 − 𝑒𝑥
= lim− 2 =1
𝑥→0
1 + 𝑒𝑥
Therefore the function is not derivable at the origin.

6.4 Differentiability and monotonicity

Monotonicity gives an idea about the behaviour of the function. A function is said to be monotonic
function if it is either increasing or decreasing in its entire domain. For example,
𝑓(𝑥) = 2𝑥 + 3has the set of all real numbers as its domain and the function is monotonically
increasing on the entire domain.

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𝑔(𝑥) = −𝑥 3 also has the set of all real numbers as its domain and the function is monotonically
decreasing on the entire domain.

We can recall that for an increasing function 𝑥1 < 𝑥2 ⇒ 𝑓(𝑥1 ) < 𝑓(𝑥2 ) and for a decreasing function
𝑥1 < 𝑥2 ⇒ 𝑓(𝑥1 ) > 𝑓(𝑥2 ). The functions which are increasing as well as decreasing in their domain
are known as non-monotonic functions. For example the absolute function, the sine function etc.

Monotonicity of a function at a point in its domain

A function is monotonically increasing at 𝑥 = 𝑎 if 𝑓(𝑎 + ℎ) > 𝑓(𝑎)and 𝑓(𝑎 − ℎ) < 𝑓(𝑎) for small ℎ >
0. From the first expression, we can write
𝑓(𝑎 + ℎ) − 𝑓(𝑎) 0
>
ℎ ℎ
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
>0
ℎ
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
lim >0
ℎ→0 ℎ
𝑓 ′ (𝑎) > 0
A function is monotonically decreasing at 𝑥 = 𝑎 if 𝑓(𝑎 + ℎ) < 𝑓(𝑎) and 𝑓(𝑎 − ℎ) > 𝑓(𝑎) for small
ℎ > 0. From the first expression, we can write
𝑓(𝑎 + ℎ) − 𝑓(𝑎) 0
<
ℎ ℎ
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
<0
ℎ
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
lim <0
ℎ→0 ℎ
𝑓 ′ (𝑎) < 0
Thus we can see a relation in differentiability and monotonicity. If a function is monotonically
increasing at 𝑥 = 𝑎, its first derivative at 𝑥 = 𝑎 has to be positive and if a function is monotonically
decreasing at 𝑥 = 𝑎, its first derivative at 𝑥 = 𝑎 has to be negative.

Monotonicity in an interval
For an increasing function in some interval if Δ𝑥 > 0 ⇔ Δ𝑦 > 0or Δ𝑥 < 0 ⇔ Δ𝑦 < 0, then the
function is said to be strictly monotonically increasing in that interval.
𝑑𝑦
i.e. if > 0 in some interval then 𝑦 is said to be a strictly increasing function in that interval.
𝑑𝑥
𝑑𝑦
Similarly, if < 0 in some interval then 𝑦 is said to be a strictly decreasing function in that interval.
𝑑𝑥
𝑑𝑦 𝑑𝑦
if ≥ 0 in some interval then 𝑦 is said to be a increasing function in that interval. Similarly, if ≤
𝑑𝑥 𝑑𝑥
0 in some interval then 𝑦 is said to be a decreasing function in that interval.

Prove that 𝑓(𝑥) = 𝑥 − sin 𝑥 is an increasing function.

Let us see how the function looks like!

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𝑓(𝑥) = 𝑥 − sin 𝑥
𝑓′(𝑥) = 1 − cos 𝑥
𝑓 ′′ (𝑥) ≥ 0 ∀ 𝑥 ∈ 𝑹
Therefore 𝑓(𝑥) is monotonically increasing ∀ 𝑥 ∈ 𝑹

Greatest and least value of a function

We can discuss the greatest and least value of a function with specific conditions under the
following three cases:
CaseI.𝑦 = 𝑓(𝑥) is strictly increasing in[𝑎, 𝑏], then 𝑓(𝑎) is the least value and 𝑓(𝑏) is the greatest
value of the function.
Case II.𝑦 = 𝑓(𝑥)is strictly decreasing in [𝑎, 𝑏], then 𝑓(𝑏) is the least value and 𝑓(𝑎) is the greatest
value of the function.
Case III.𝑦 = 𝑓(𝑥)is non-monotonic in [𝑎, 𝑏] and is continuous, then the greatest and least value of
𝑓(𝑥) in [𝑎, 𝑏] are those values where 𝑓 ′ (𝑥) = 0 or it does not exist or at the extreme values.

Find the interval in which the function 𝑓(𝑥) = 2𝑥 2 − ln |𝑥| is (i) decreasing (ii) increasing.
𝑓(𝑥) = 2𝑥 2 − ln|𝑥|
1 4𝑥 2 − 1
𝑓 ′ (𝑥) = 4𝑥 − =
x 𝑥
Domain of the function is(0, ∞). Therefore the denominator of 𝑓 ′ (𝑥) is always positive and
numerator has all the power to decide.
For 𝑓(𝑥) to be decreasing
𝑓 ′ (𝑥) ≤ 0
4𝑥 2 − 1 ≤ 0
1
𝑥2 ≤
4
1
|𝑥| ≤
2
1 1
− ≤𝑥≤
2 2
1
𝑥 ∈ (0, ]
2
For 𝑓(𝑥) to be increasing

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Notes Calculus
𝑓 ′ (𝑥) ≥ 0
4𝑥 2 − 1 ≥ 0
1
𝑥2 ≥
4
1
|𝑥| ≥
2
1 1
𝑥 ≥ 𝑜𝑟 𝑥 ≤ −
2 2
1
𝑥 ∈ [ , ∞)
2
We can verify this by drawing the graph of the function and can observe the following:

1
• From onwards the function is increasing
2
1
• In (0, ] the function is decreasing.
2
1
• At the function is having a minimum value.
2

Summary
In this unit we have learnt the basic definition of differentiability and its geometric and kinematic
interpretations. We have learnt some results related to differentiability with continuity and
monotonicity. The following are the main point:
𝑓(𝑐+ℎ)−𝑓(𝑐)
• To check the derivability, we need to check if the limit lim exists or not.
ℎ→0 ℎ
• If f is finitely derivable at c, then f is also continuous at c.
• If a function is monotonically increasing at 𝑥 = 𝑎, its first derivative at 𝑥 = 𝑎 has to be
positive and if a function is monotonically decreasing at 𝑥 = 𝑎, its first derivative at 𝑥 = 𝑎
has to be negative.

Key Words
• Differentiability
• Derivability
• Differentiability and continuity
• Differentiability and monotonicity

Self Assessment
1. Which of the following does not lead to the idea of differentiability?
A. instantaneous rate of change

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 Unit 06: Differentiability of a Real Valued Function Notes
B. average rate of change
C. slope of the function
D. local rate of change

2. A function 𝑓 is said to be derivable in [𝑎, 𝑏] if

A. 𝑓 is finitely derivable at every point of [𝑎, 𝑏]
B. 𝑓 is infinitely derivable at every point of [𝑎, 𝑏]
C. 𝑓 is finitely derivable at some points of [𝑎, 𝑏]
D. 𝑓 is infinitely derivable at some points of [𝑎, 𝑏

3. The derivative of the function √𝑥 + 2 is

1
A.
𝑥+2
1
B.
2(𝑥+2)
1
C.
2√𝑥+2
1
D.
2√𝑥

4. A function 𝑓 is said to be derivable at point 𝑐 if

𝑓(𝑐+ℎ)−𝑓(𝑐)
A. lim− exists
ℎ→0 ℎ
𝑓(𝑐+ℎ)−𝑓(𝑐)
B. lim+ exists
ℎ→0 ℎ
𝑓(𝑐+ℎ)+𝑓(𝑐)
C. lim exists
ℎ→0 ℎ
𝑓(𝑐+ℎ)−𝑓(𝑐)
D. lim exists
ℎ→0 ℎ

5. The derivative of 𝑐𝑜𝑠𝑒𝑐 𝑥 w.r.t. 𝑥 is

A. 𝑐𝑜𝑠𝑒𝑐𝑥 cot 𝑥
B. −𝑐𝑜𝑠𝑒𝑐𝑥 cot 𝑥
C. −𝑐𝑜𝑠𝑒𝑐 2 𝑥 cot 𝑥
D. −𝑐𝑜𝑠𝑒𝑐𝑥 cot 2 𝑥

6. The derivative of 𝑙𝑜𝑔𝑎 𝑥 w.r.t. 𝑥 is

1
A.
𝑥
1
B.
𝑥𝑙𝑜𝑔𝑎𝑥
1
C.
𝑥𝑙𝑜𝑔𝑒 𝑎
D. none of these

7. The derivative of 𝑠𝑖𝑛−1 𝑥 w.r.t. 𝑥 is

1
A.
1−𝑥 2
1
B.
√1−𝑥 2
1
C.
1+𝑥 2
1
D.
√1+𝑥 2

8. The derivative of 𝑐𝑜𝑡 −1 𝑥 w.r.t. 𝑥 is

1
A.
1−𝑥 2
1
B.
√1−𝑥 2
1
C.
1+𝑥 2

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Notes Calculus
D. none of these

9. The derivative of 𝑡𝑎𝑛ℎ−1 𝑥 w.r.t. 𝑥 is

1
A.
1−𝑥 2
1
B.
√1−𝑥 2
1
C.
1+𝑥 2
D. none of these

10. The derivative of 𝑐𝑜𝑠ℎ 𝑥 w.r.t. 𝑥 is

A. sinh 𝑥
B. − sinh 𝑥
C. − sinh2 𝑥
D. sinh2 𝑥

11. Which of the following function is continuous and not differentiable in its domain?
A. 𝑥 2
B. √𝑥
C. |𝑥|
1
D.
𝑥

12. The function 𝑥|𝑥| is

A. derivable at origin
B. continuous at origin
C. both derivable and continuous at origin
D. none of these

13. If f is continuous at c, then f is also finitely derivable at c.

A. True
B. False

14. A function is said to be monotonic function if

A. it is increasing in its entire domain
B. it is decreasing in its entire domain
C. it is either increasing or decreasing in its entire domain
D. none of these

15. Which of the following suggests that the function is strictly decreasing?
A. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) < 𝑓(𝑥2 )
B. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) > 𝑓(𝑥2 )
C. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) ≤ 𝑓(𝑥2 )
D. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) ≥ 𝑓(𝑥2 )

16. Which of the following suggests that the function is decreasing?

A. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) < 𝑓(𝑥2 )
B. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) > 𝑓(𝑥2 )
C. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) ≤ 𝑓(𝑥2 )
D. 𝑥1 < 𝑥2 implies 𝑓(𝑥1 ) ≥ 𝑓(𝑥2 )

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 Unit 06: Differentiability of a Real Valued Function Notes

17. The functions which are increasing as well as decreasing in their domain are known as
A. increasing functions
B. decreasing functions
C. monotonic functions
D. non monotonic functions

18. The function 2𝑥 2 − 𝑙𝑜𝑔 𝑥 is decreasing in the interval

1 1
A. [− , ]
2 2
1
B. [0, ]
2
1
C. (0, )
2
1
D. (0, ]
2

Answers for Self Assessment

1. B 2. A 3. C 4. B 5. D

6. C 7. B 8. D 9. A 10. A

11. C 12. C 13. B 14. C 15. B

16. D 17. D 18. D

Review Questions
1. Find the derivative of 𝑓(𝑥) = 2𝑎𝑥 + 𝑏 using first principle.
1
2. Find the derivative of 𝑓(𝑥) = using first principle.
𝑥 2 +3
3. Discuss the differentiability of the function (𝑥) = |𝑥 − 2| + |𝑥| + |𝑥 + 2| .
4. Find the slope of the tangents to the parabola 𝑦 = 𝑥 2 at points (2, 4) and (-1, 1).
5. Find the interval in which the function 𝑓(𝑥) = 3𝑥 2 − ln |𝑥| is (i) decreasing (ii) increasing.
6. Find the interval in which the function 𝑓(𝑥) = log 𝑥 + 𝑥 is (i) decreasing (ii) increasing.
1 1
𝑥(𝑒 − 𝑥 −𝑒 𝑥 )
7. Examine 𝑓(𝑥) = { 1 1 𝑥 ≠ 0 for the continuity and differentiability at origin.
𝑒 − 𝑥 +𝑒 𝑥
1 𝑥=0
8. Find the interval in which the function 𝑓(𝑥) = 𝑥 − cos 𝑥 is (i) decreasing (ii) increasing.
9. Find the derivative of hyperbolic sine function using ab initio method.
10. Design a function which is increasing on some part of the domain and decreasing on
other. Then discuss the differentiability and continuity of that function.

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Notes Calculus

Further Readings
George B. Thomas Jr., Joel Hass, Christopher Heil& Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens& Stephan Davis (2016). Calculus (10th edition). Wiley India.
A Geometric Interpretation of the Derivatives. (2020, November 17). Retrieved May 4,
2021, from https://math.libretexts.org/@go/page/25429

Web Links
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins
_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits_Continuity_and_Differenti
ability

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 Notes
Unit 07: Differentiability of a Real Valued Function
Dr. Pratiksha, Lovely Professional University

Unit 7: Differentiability of a Real Valued Function

CONTENTS
Objectives
Introduction
7.1 Chain Rule of Differentiation
7.2 Extreme Value Theorem
7.3 Darboux's Theorem
7.4 Rolle’s Theorem
Summary
Key Words
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will
• learn about the derivative of the function of function
• explore the property of a differentiable function called Darboux’s theorem
• learn to apply the Rolle’s theorem

Introduction
If the function is made up of functions called composed functions or a composite function, then
what to do in case, if we are interested in the derivative of a composed function! This question will
be answered in this unit. The derivative of function of function is popularly known as the chain rule
of differentiation.
Let 𝑓 and 𝜙 be two derivable functions such that 𝑦 = 𝑓(𝑢)and 𝑢 = 𝜙(𝑥). Clearly you can see that 𝑦
is a function of 𝑢 and 𝑢 is a function of 𝑥 and ultimately 𝑦 is a function of 𝑥.

𝑦 𝑢 𝑥

The range of 𝜙 must be a subset of the domain of 𝑓, then only we would be able to write 𝑦 =
𝑓(𝜙(𝑥)) which is also called the composite function. Moreover we know that
(𝑓𝑜𝜙)(𝑥) = 𝑓(𝜙(𝑥))

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Notes
Calculus

7.1 Chain Rule of Differentiation

The chain rule is a rule for differentiating compositions of functions.
𝑑 𝑑 𝑑
(𝑓(𝜙(𝑥)) = 𝑓(𝜙(𝑥)). 𝜙(𝑥)
𝑑𝑥 𝑑(𝜙(𝑥)) 𝑑𝑥

Or it can be put simply as

𝑑
(𝑓(𝜙(𝑥)) = 𝑓 ′ (𝜙(𝑥))𝜙′(𝑥)
𝑑𝑥
However, we rarely use this formal approach when applying the chain rule to specific problems.
Instead, we take an intuitive approach. For example, it is sometimes easier to think of the functions
𝑓 and 𝜙 as ``layers'' of a problem. Function 𝑓 is the ``outer layer'' and function 𝜙 is the ``inner
layer.'' Thus, the chain rule tells us to first differentiate the outer layer, leaving the inner layer
unchanged (the term 𝑓 ′ (𝜙(𝑥)) , then differentiate the inner layer (the term 𝜙′(𝑥) ) .
The chain rule provides us a technique for finding the derivative of composite functions, with the
number of functions that make up the composition determining how many differentiation steps are
necessary. For example, if a composite function 𝑓(𝑥) is defined as
𝑓(𝑥) = (𝑔𝑜ℎ)(𝑥) = 𝑔(ℎ(𝑥))
Then 𝑓 ′ (𝑥) = 𝑔′(ℎ(𝑥). ℎ′(𝑥)
If the function is defined as 𝑓(𝑥) = (𝑔𝑜ℎ𝑜𝑘)(𝑥) = 𝑔(ℎ(𝑘(𝑥)))

Then 𝑓 ′ (𝑥) = 𝑔′(ℎ(𝑘(𝑥)). ℎ′ (𝑘(𝑥)). 𝑘′(𝑥)

Let us consider 𝑓 and 𝜙 be two derivable functions such that 𝑦 = 𝑓(𝑢)and 𝑢 = 𝜙(𝑥). Let
Δ𝑥be the change in 𝑥
Δ𝑢be the change in 𝑢
Δ𝑦be the change in 𝑦.
Then we can write
Δ𝑦 Δ𝑦 Δ𝑢
= .
Δ𝑥 Δ𝑢 Δ𝑥
Δ𝑦 Δ𝑦 Δ𝑢
lim = lim ( . )
Δ𝑥→0 Δ𝑥 Δ𝑥→0 Δ𝑢 Δ𝑥

Δ𝑦 Δ𝑦 Δ𝑢
lim = lim . lim
Δ𝑥→0 Δ𝑥 Δ𝑥→0 Δ𝑢 Δ𝑥→0 Δ𝑥

𝑑𝑦 𝑑𝑦 𝑑𝑢
= .
𝑑𝑥 𝑑𝑢 𝑑𝑥
Let us understand the rule better with the following examples:

Find the derivative of the function √1 + 𝑥 2 .

You can see that the function is a composition of the polynomial function and the square root
function. So we can write this as

𝑢 = 1 + 𝑥 2 , 𝑦 = √𝑢; then𝑦 = √1 + 𝑥 2
𝑑𝑢
Now = 2𝑥
𝑑𝑥

𝑑𝑦 1 −1 1 1
= 𝑢 2 = (1 + 𝑥 2 )−2
𝑑𝑢 2 2
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑥
Therefore, = . = .
𝑑𝑥 𝑑𝑢 𝑑𝑥 √1+𝑥 2

We can look at the same problem by layers’ point of view. The square root is the outer layer , it has
to be dealt with first and then the polynomial as the inner layer will be considered. We can write it
as

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Unit 07: Differentiability of a Real Valued Function
𝑑 𝑑√1 + 𝑥 2 𝑑(1 + 𝑥 2 )
(√1 + 𝑥 2 ) = .
𝑑𝑥 𝑑(1 + 𝑥 2 ) 𝑑𝑥
1 1
= (1 + 𝑥 2 )−2 . 2𝑥
2
𝑥
=
√1 + 𝑥 2

1+𝑥
Find the derivative of the function √ .
1−𝑥
1
1+𝑥
Let = 𝑢, then 𝑦 = √𝑢 = 𝑢2
1−𝑥
1
𝑑𝑦 1 −1 1 1 + 𝑥 −2
= 𝑢 2 = .( )
𝑑𝑢 2 2 1−𝑥
𝑑𝑢 2
=
𝑑𝑥 (1 − 𝑥)2
1 1
𝑑𝑦 1 1 + 𝑥 −2 2 (1 + 𝑥)−2
= ( ) . = 3
𝑑𝑥 2 1 − 𝑥 (1 − 𝑥)2 (1 − 𝑥)2

Find the derivative of the function log(cosh 𝑥).

𝑑𝑦 𝑑 log(cosh 𝑥) 𝑑(𝑐𝑜𝑠ℎ𝑥)
= .
𝑑𝑥 𝑑(𝑐𝑜𝑠ℎ𝑥) 𝑑𝑥
1
= . 𝑠𝑖𝑛ℎ𝑥 = tanh 𝑥
𝑐𝑜𝑠ℎ𝑥

Find the derivative of the function sinh−1 𝑥.

Let 𝑦 = sinh−1 𝑥
⇒ 𝑥 = sinh 𝑦

𝑑𝑥
= cosh 𝑦
𝑑𝑦
𝑑𝑦 1
⇒ =
𝑑𝑥 cosh 𝑦
1
=±
√(1 + sinh2 𝑦)
1
=±
√1 + 𝑥 2
The sign of the radical must be same as that of cosh 𝑦.
𝑑𝑦 1
Therefore, =
𝑑𝑥 √1+𝑥 2

−1 𝑥
Find the derivative of the function 𝑒 𝑠𝑖𝑛ℎ .
Let 𝑦 = sinh−1 𝑥
−1
𝑑𝑦 𝑑(𝑒 𝑠𝑖𝑛ℎ 𝑥 ) 𝑑(sinh−1 𝑥)
= .
𝑑𝑥 𝑑(sinh−1 𝑥) 𝑑𝑥
−1 𝑥 1
= 𝑒 sinh .
√1 + 𝑥 2

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Notes
Calculus
7.2 Extreme Value Theorem
An major application of critical points or saddle points is in determining possible maximum and
minimum values of a function on certain intervals. The Extreme Value Theorem guarantees both a
maximum and minimum value for a function under certain conditions. It states the following:
If a function f(x) is continuous on a closed interval [a, b], then f(x) has both a maximum and
minimum value on [a, b].
The steps for applying the extreme value theorem are as follows:
1. Establish that the function is continuous on the closed interval.
2. Determine all critical points in the given interval.
3. Evaluate the function at these critical points and at the endpoints of the interval.
4. Look for the largest and the smallest values of the function.
The largest function value from the previous step is the maximum value, and the smallest function
value is the minimum value of the function on the given interval.

Find the maximum and minimum values of 𝑓(𝑥) = sin 𝑥 + cos 𝑥 on [0, 2𝜋].
The function is continuous on [0, 2𝜋].
𝑓 ′ (𝑥) = cos 𝑥 − sin 𝑥
𝜋 5𝜋
The critical points are ( , √2)and ( , −√2). The function values at the end pointsof the given
4 4
interval are 𝑓(0) = 1and 𝑓(2𝜋) = 1.

Thus, we can see that the maximum value of the function is √2

and the minimum value is −√2.

Find the maximum and minimum values of 𝑓(𝑥) = 𝑥 4 −

3𝑥 − 1 on [−2, 2].
3

The function is a polynomial, therefore is continuous on [−2, 2].

Its derivative is given as 𝑓 ′ (𝑥) = 4𝑥 3 − 9𝑥 2
For critical points, put 𝑓 ′ (𝑥) = 0
or4𝑥 3 − 9𝑥 2 = 0
9
⇒ 𝑥 = 0, .
4
9
Clearly 𝑥 = does not belong to the interval [−2, 2]. The only
4
critical point occurs at 𝑥 = 0. which is (0, −1).
The function values at the endpoints of the interval are f(2)=−9 and f(−2)=39; therefore, the
maximum function value is 39 at x = −2, and the minimum function value is −9 at x = 2.
Note the importance of the closed interval in determining the values to consider for critical points.

7.3 Darboux's Theorem

If f is differentiable on the closed interval [a, b] and r is any number between f ’(a) and f’ (b), then
there exists a number c in the open interval (a, b) such that f ‘ (c) = r.

Proof
Consider the function
ℎ(𝑥) = 𝑓(𝑥) − (𝑓(𝑏) + 𝑟(𝑥 − 𝑏))
Because 𝑓(𝑥) is differentiable, it is definitely continuous.
𝑓(𝑏) + 𝑟(𝑥 − 𝑏) is also continuous and differentiable.
∴ ℎ(𝑥)is continuous and differentiable on [𝑎, 𝑏].

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Unit 07: Differentiability of a Real Valued Function
By the extreme value theorem, there exists 𝑐 ∈ (𝑎, 𝑏) where ℎ(𝑥) has an extreme value . At this point
ℎ′ (𝑐) = 0
We have ℎ′ (𝑥) = 𝑓 ′ (𝑥) − 𝑟
So 𝑓 ′ (𝑐) − 𝑟 = 0
⇒ 𝑓 ′ (𝑐) = 𝑟

Thus if f is differentiable on the closed interval [a, b] and r is any number between f ’(a) and f ’ (b),
then there exists a number c in the open interval (a, b) such that f ‘ (c) = r.

Jean Gaston Darboux was a French mathematician who lived from 1842 to 1917. Of his
several important theorems the one we just studied says that the derivative of a function has the
Intermediate Value Theorem property – that is, the derivative takes on all the values between the
values of the derivative at the endpoints of the interval under consideration.
Another interesting aspect of Darboux’s Theorem is that there is no requirement that the derivative
f ‘(x) be continuous!
1
The common example of such a function is𝑓(𝑥) = 𝑥 2 sin when 𝑥 ≠ 0 and 𝑓(𝑥) = 0 when 𝑥 = 0
𝑥
1 1
With𝑓 ′ (𝑥) = 2𝑥 sin − cos , 𝑥 ≠ 0
𝑥 𝑥

This function is differentiable and hence continuous. There is an oscillating discontinuity at the
origin. The derivative is not continuous at the origin. Yet, every interval containing the origin as an
interior point meets the conditions of Darboux’s Theorem, so the derivative while not being
continuous has the intermediate value property.

7.4 Rolle’s Theorem

If f(x) is continuous an [a,b] and differentiable on (a,b)

and if f(a) = f(b) then there is some c in the interval (a,b)
such that f '(c) = 0.

Proof

In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. Hence
by the Intermediate Value Theorem it achieves a maximum and a minimum on [a,b]. Either one of
these occurs at a point c with a < c < b,
Since f(x) is differentiable on (a,b) and c is an extremum we then conclude that f '(c) = 0.
or both the maximum and minimum occur at endpoints.
Since f(a) = f(b), this means that the function is never larger or smaller than f(a). In other words, the
function f(x) is constant on the interval [a,b] and its derivative is therefore 0 at every point in (a,b).
Hence proved
Geometric interpretation

There is a point c on the interval (a,b)where the tangent to the graph of the function is parallel to
the x-axis.

This property was known in the 12th century in ancient India. The outstanding Indian
astronomer and mathematician Bhaskara II mentioned it in his writings.
For instance, consider𝑓(𝑥) = |𝑥| (where |𝑥|is the absolute value of 𝑥 on the closed interval [-1,1].
This function does not have derivative at= 0. Though f(x) is continuous on the closed interval [-1,1]
there is no point inside the interval (−1,1)at which the derivative is equal to zero. The Rolle’s
Theorem fails here because𝑓(𝑥) is not differentiable over the whole interval(−1,1).

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Notes
Calculus
Physical interpretation
Rolle’s Theorem has a clear physical meaning. Suppose that a body moves along a straight line, and
after a certain period of time returns to the starting point. Then, in this period of time there is a
moment, in which the instantaneous velocity of the body is equal to zero.

Verify the Rolle’s Theorem for 𝑓(𝑥) = 𝑥 2 in [−1,1]

The function is a polynomial, therefore it is continuous in [−1,1].
The function is differentiable in (−1,1). (You can verify it by ab initio definition or but just checking
that its derivative by the usual rules of differential calculus, exists in (−1,1).
And thirdly 𝑓(1) = 𝑓(−1)
All conditions are getting fulfilled therefore in the interval(−1,1), there must exist at least a point
such that the derivative of the function at that point is zero.
i.e.𝑓 ′ (𝑐) = 0
⇒ 2𝑐 = 0

⇒ 𝑐=0

And 0 ∈ (−1,1)
Thus the Rolle’s Theorem gets verified.

𝑥
Verify Rolle’s theorem for 𝑓(𝑥) = 𝑥(𝑥 + 3)𝑒 −2 in [−3,0].
The function is a product of continuous functions, therefore it is continuous in [−3,0].
The function is differentiable in (−3,0). (solve for the derivative of the function and check if it exists
in (−3,0), It will be!)
𝑓(−3) = 𝑓(0)
Therefore there will exist a point 𝑐 , such that𝑓 ′ (𝑐) = 0.
Or 𝑐 2 − 𝑐 − 6 = 0
Or 𝑐 = −2, 3
Thus we got at least a point −2 ∈ (−3,0).
Thus Rolle’s theorem gets verified for the given function in the given interval.

Summary
This unit is an extension of the differentiability to a function of a function. We have understood and
learnt the formulae of the derivatives of elementary functions already.
• In order to differentiate a composite function, of course those formulae will not be
applicable directly. We need to use the chain rule.
𝑑
• (𝑓(𝜙(𝑥)) = 𝑓 ′ (𝜙(𝑥))𝜙′(𝑥)
𝑑𝑥
• A significant result for a differentiable function on a closed interval, given as,
If f is differentiable on the closed interval [a, b] and r is any number between f ’(a) and
f ’ (b), then there exists a number c in the open interval (a, b) such that f ‘ (c) = r.
• Rolle’s theorem states that ‘If f(x) is continuous an [a,b] and differentiable on (a,b) and if
f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0.’

Key Words
chain rule, derivative of a function of function, extreme value theorem, Darboux’s theorem, Rolle’s
theorem

Self Assessment
1. The derivative of the function √2 + 𝑥 2 is
𝑥
A. 2
2√2+𝑥
𝑥
B.
√2+2𝑥 2

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 Notes
Unit 07: Differentiability of a Real Valued Function
2𝑥
C.
√2+𝑥 2
𝑥
D.
√2+𝑥 2

𝑑
2. (𝑠𝑖𝑛(𝑐𝑜𝑠𝑥)) =
𝑑𝑥
A. sin 𝑥 cos (cos 𝑥)
B. cos (sin 𝑥)
C. −sin 𝑥 cos (cos 𝑥)
D. sin 𝑥 cos (sin 𝑥)

3. If f is differentiable on the closed interval [a, b] and r is any number between f ’ (a) and
f ’(b), then there exists a number c in the open interval (a, b) such that f ’(c) = r. This
statement is of
A. intermediate value theorem
B. mean value theorem
C. Rolle’s theorem
D. Darboux’s theorem

−1 𝑥
𝑒 𝑠𝑖𝑛ℎ
4. is the derivative of the function
√1+𝑥 2
−1
𝑒 𝑠𝑖𝑛ℎ 𝑥
A. x
√1+𝑥 2
−1
𝑒 𝑠𝑖𝑛ℎ 𝑥
B.
√1+𝑥 2
−1
C. 𝑒 𝑠𝑖𝑛ℎ 𝑥 ans
−1
D. 𝑒 𝑐𝑜𝑠ℎ 𝑥

5. The derivative of 𝑙𝑜𝑔(𝑐𝑜𝑠ℎ 𝑥) w.r.t. 𝑥 is

A. log (sin 𝑥)
B. log (sinh 𝑥)
C. 𝑐𝑜𝑠𝑒𝑐ℎ 𝑥
D. tanh 𝑥

𝑑 𝑠𝑖𝑛𝑥
6. 𝑒 =
𝑑𝑥
A. 𝑒 𝑠𝑖𝑛𝑥
cos 𝑥
B. cos (sin 𝑥)
C. − 𝑒 𝑠𝑖𝑛𝑥 cos 𝑥
D. sin 𝑥 cos (sin 𝑥)

7. If a function 𝑓 is
I. continuous on [𝑎, 𝑏]
II. derivable on (𝑎, 𝑏)
III. 𝑓(𝑎) = 𝑓(𝑏)
IV. then there exists one value 𝑐 ∈ (𝑎, 𝑏) such that 𝑓 ′ (𝑐) = 0
Which of the following are correct for 𝑓 to satisfy the Rolle’s Theorem?
A. I, II and III
B. I, II and IV
C. II, III and IV
D. III and IV

8. The function 𝑓(𝑥) = 𝑥 2 in [−1, 2] satisfies the Rolles’s theorem.

A. True
B. False

𝑥
9. The function 𝑓(𝑥) = 𝑥(𝑥 + 3)𝑒 −2 in [−1, 1] satisfies the Rolle’s Theorem.
A. True
B. False

10. For all the second degree polynomials 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑘, it is seen that the Rolles’ point is
at 𝑐 = 0. Also the value of 𝑘 is zero. Then what is the value of 𝑏?

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Calculus
A. 0
B. 1
C. -1
D. 56

11. If 𝑓 is continuous function on the closed interval [𝑎, 𝑏], and 𝑁 is a number between 𝑓(𝑎)
and 𝑓(𝑏), then there is 𝑐 ∈ [𝑎, 𝑏]such that 𝑓(𝑐) = 𝑁 is:
A. The Intermediate Value Theorem
B. The Mean Value Theorem
C. Rolle's Theorem
D. The Extreme Value Theorem

12. According to Rolle's theorem, for a differentiable function f(x), if the start point f(a) and
the end point f(b) equal 0 then:
A. Rolle's Theorem does not apply.
B. Somewhere between f(a) and f(b) the instantaneous rate of change must be 0.
C. Somewhere between f(a) and f(b) the function must equal 0.
D. The function is flat.

Answer for Self Assessment

1. D 2. C 3. D 4. C 5. D

6. A 7. A 8. B 9. B 10. A

11. A 12. B

Review Questions
1. Find the derivative of the function 𝑓(𝑥) = sin(√𝑥 2 − 5)
√1−𝑥
2. Find the derivative of the function 𝑓(𝑥) =
1+𝑥
3. Find the derivative of the function 𝑓(𝑥) = sin ℎ(√𝑥 2 + 5)
4. Find the derivative of the function 𝑓(𝑥) = log (cosh 𝑒 𝑥 )
𝑥
5. Find the derivative of the function 𝑓(𝑥) = 2(cosh 𝑒 )
6. Find the derivative of the function 𝑓(𝑥) = tan 𝑥−1

7. Design a function which satisfies the Darboux’s theorem in certain interval.

8. State and prove the Rolle’s theorem.
9. Learn more about Michel Rolle!
10. Compare the Rolles’ theorem with the Darboux’s theorem.
11. Discuss the applicability of Rolle’s theorem to the function
𝑥 2 + 1, 0 ≤ 𝑥 ≤ 1
𝑓(𝑥) = {
3 − 𝑥, 1 < 𝑥 ≤ 2

Further Reading
George B. Thomas Jr., Joel Hass, Christopher Heil & Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens& Stephan Davis (2016).Calculus (10th edition).Wiley India.

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/Chain
Rule.html
https://www.cliffsnotes.com/study-guides/calculus/calculus/the-derivative/chain-rule
https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-
derivative/extreme-value-theorem
https://www.math24.net/rolles-theorem

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Dr. Pratiksha, Lovely Professional University Unit 08: Mean Value Theorems

Unit 08: Mean Value Theorems

CONTENTS
Objectives
Introduction
8.1 Lagrange’s Mean Value Theorem
8.2 Alternate form of Lagrange’s Mean Value Theorem
8.3 Geometric and Physical Interpretation of Lagrange’s Mean Value Theorem
8.4 Cauchy’s Mean Value Theorem
8.5 Alternate form of Lagrange’s Mean Value Theorem
Summary
Key Words
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will
• learn about the properties of a differentiable function
• understand the basics of Lagrange’s mean value theorem
• be able to use Cauchy’s mean value theorem
• be able to interpret the mean value theorems geometrically

Introduction
If a function is appropriately differentiable and continuous then it can lead to much more
information about the nature and behavior of the function. In this chapter we will learn the more
general form of the Rolle’s Theorem and then the general form of the Lagrange mean value
theorem with their physical interpretations.

8.1 Lagrange’s Mean Value Theorem

This theorem is also called the First Mean Value Theorem and allows to express the increment of a
function on an interval through the value of the derivative at an intermediate point of the segment.
Lagrange’s mean value theorem (MVT) states that if a function𝑓(𝑥) is continuous on a closed
interval[𝑎, 𝑏]and differentiable on the open interval(𝑎, 𝑏), then there is at least one point𝑥 = 𝑐 on
𝑓(𝑏)−𝑓(𝑎)
this interval, such that = 𝑓′(𝑐).
𝑏−𝑎

Proof
Define a new function
𝜙(𝑥) = 𝑓(𝑥) + 𝐴𝑥, 𝑥 ∈ [𝑎, 𝑏]
We choose a number 𝐴 such that the condition 𝜙(𝑎) = 𝜙(𝑏) is satisfied. Then
𝑓(𝑎) + 𝐴𝑎 = 𝑓(𝑏) + 𝐴𝑏
𝑓(𝑏) − 𝑓(𝑎)
𝐴=−
𝑏−𝑎
As a result, we have

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𝑓(𝑏) − 𝑓(𝑎)
𝜙(𝑥) = 𝑓(𝑥) + (− )𝑥
𝑏−𝑎
The function 𝜙(𝑥) is continuous on the closed interval[𝑎, 𝑏], differentiable on the open interval
(𝑎, 𝑏) and takes equal values at the endpoints of the interval. Therefore, it satisfies all the conditions
of the Rolle’s Theorem. Then there is at least a point 𝑐 in the interval (𝑎, 𝑏) such that
𝜙 ′ (𝑐) = 0
It follows that
𝑓 ′ (𝑐) + 𝐴 = 0
or
𝑓 ′ (𝑐) = −𝐴
or
𝑓(𝑏)−𝑓(𝑎)
𝑓 ′ (𝑐) =
𝑏−𝑎

Joseph Louis Lagrange, the greatest mathematician of

the eighteenth century, was born at Turin on January 25,
1736, and died at Paris on April 10, 1813. His father, who
had charge of the Sardinian military chest, was of good
social position and wealthy, but before his son grew up he
had lost most of his property in speculations, and young
Lagrange had to rely for his position on his own abilities.
He was educated at the college of Turin, but it was not
until he was seventeen that he showed any taste for
mathematics - his interest in the subject being first excited
by a memoir by Halley across which he came by accident.
Alone and unaided he threw himself into mathematical
studies; at the end of a year's incessant toil he was already an accomplished mathematician, and
was made a lecturer in the artillery school.

8.2 Alternate form of Lagrange’s Mean Value Theorem

If a function 𝑓(𝑥) is continuous on a closed interval [𝑎, 𝑎 + ℎ]and differentiable on the open interval
(𝑎, 𝑎 + ℎ), then there is at least one 𝜃 ∈ (0,1) such that
𝑓(𝑎+ℎ)−𝑓(𝑎)
= 𝑓′(𝑎 + 𝜃ℎ).
ℎ

Let 𝑏 − 𝑎 = ℎ = length of the interval [𝑎, 𝑏]

Therefore [𝑎, 𝑏] can be written as [𝑎, 𝑎 + ℎ]
Also 𝑎 < 𝑐 < 𝑎 + ℎ
Therefore 𝑐 can be written as 𝑎 + 𝜃ℎ where 𝜃 ∈ (0,1).
So, the expression
𝑓(𝑏) − 𝑓(𝑎)
𝑓 ′ (𝑐) = −
𝑏−𝑎
becomes
𝑓(𝑎 + ℎ) − 𝑓(𝑎) = ℎ 𝑓 ′ (𝑎 + 𝜃ℎ)

or
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ 𝑓 ′ (𝑎 + 𝜃ℎ), 𝜃 ∈ (0,1)

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Unit 08: Mean Value Theorems
8.3 Geometric and Physical Interpretation of Lagrange’s Mean Value
Theorem
Geometric interpretation
Lagrange’s mean value theorem has a simple geometrical meaning. The chord passing through the
points of the graph corresponding to the ends of the segment 𝑎 and 𝑏 has the slope equal to
𝑓(𝑏) − 𝑓(𝑎)
𝑏−𝑎
Then there is a point 𝑥 = 𝑐inside the interval[𝑎, 𝑏] where the tangent to the graph is parallel to the
chord.
Physical interpretation
The mean value theorem has also a clear physical interpretation. If we assume that𝑓(𝑡) represents
the position of a body moving along a line, depending on the time𝑡, then the ratio
is the average velocity of the body in the period of time𝑏 − 𝑎. Since𝑓′(𝑡)is the instantaneous
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
velocity, this theorem means that there exists a moment of time 𝑐 at which the instantaneous
velocity is equal to the average velocity.
Lagrange’s mean value theorem has many applications in mathematical analysis, computational
mathematics and other fields. Let us further note two remarkable corollaries.
Corollary I
In a particular case when the values of the function𝑓(𝑥) at the endpoints of the segment[𝑎, 𝑏]are
equal, i.e.𝑓(𝑎) = 𝑓(𝑏)the mean value theorem implies that there is a point𝑐 ∈ (𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎)
𝑓 ′ (𝑐) = =0
𝑏−𝑎
and that is the Rolle’s theorem, which can hence be considered as a special case of Lagrange’s mean
value theorem.
Corollary II
If the derivative 𝑓′(𝑥) is zero at all points of the interval[𝑎, 𝑏] then the function𝑓(𝑥) is constant on
this interval.
For any two points𝑥1 and 𝑥2 in the interval[𝑎, 𝑏], there exists a point 𝑐 ∈ (𝑎, 𝑏)such that
𝑓(𝑥2 ) − 𝑓(𝑥1 ) = 𝑓 ′ (𝑐)(𝑥2 − 𝑥1 ) = 0
And this results in
𝑓(𝑥1 ) = 𝑓(𝑥2 )

If 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3), 𝑥 ∈ [0,4], find 𝑐 such that the average rate of change of 𝑓(𝑥)
is equal to the derivative of f(x) at 𝑐.
Here,
𝑓(0) = −6
𝑓(4) = 6
𝑓(4)−𝑓(0)
So, =3
4−0

Also 𝑓 ′ (𝑥) = 3𝑥 2 − 12 𝑥 + 11
According to the statement,
𝑓(4) − 𝑓(0)
= 𝑓′(𝑐)
4−0
3𝑐 2 − 12𝑐 + 11 − 3 = 0
6 ± 2√3
⇒ 𝑐=
8

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Verify the mean value theorem for 𝑓(𝑥) = 𝑥 3 in [𝑎, 𝑏]

The function is a polynomial, therefore continuous in [𝑎, 𝑏]
𝑓 ′ (𝑥) = 3𝑥 2 exists in (a,b), therefore the function is differentiable in (a,b).
Since the function is satisfying the requirements of Lagrange’s mean value theorem, there must
exist a 𝑐 ∈ (𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎)
𝑓 ′ (𝑐) =
𝑏−𝑎
𝑏3 − 𝑎3
3𝑐 2 =
𝑏−𝑎
𝑏 2 + 𝑎𝑏 + 𝑎2 = 3𝑐 2
𝑏 2 + 𝑎𝑏 + 𝑎2
𝑐2 =
3

𝑏 2 + 𝑎𝑏 + 𝑎2
𝑐 = ±√
3

Prove that for any quadratic function 𝑝𝑥 2 + 𝑞𝑥 + 𝑟, the value of 𝜃 in Lagrange’s theorem is
1
always , for any 𝑝, 𝑞, 𝑟, 𝑎, ℎ.
2

Let 𝑓(𝑥) = 𝑝𝑥 2 + 𝑞𝑥 + 𝑟, 𝑥 ∈ [𝑎, 𝑎 + ℎ]

Clearly, the given function is continuous in [𝑎, 𝑎 + ℎ]
and derivable in (𝑎, 𝑎 + ℎ)
Therefore, there exists a 𝜃 in (0,1) such that
𝑓(𝑎 + ℎ) − 𝑓(𝑎) = ℎ𝑓 ′ (𝑎 + 𝜃ℎ)
Substituting and simplifying,
𝑝ℎ2 = 2𝑝𝜃ℎ2
1
⇒ 𝜃= ∈ (0,1)
2
Hence proved

8.4 Cauchy’s Mean Value Theorem

If two functions 𝑓(𝑥) and 𝐹(𝑥) are continuous on an interval [𝑎, 𝑏], differentiable on (𝑎, 𝑏) and F’(x)
is non zero for all 𝑥 ∈ (𝑎, 𝑏), then there exists a point 𝑐 ∈ (𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎) 𝑓 ′ (𝑐)
=
𝐹(𝑏) − 𝐹(𝑎) 𝐹 ′ (𝑐)
This theorem is also known as generalized Lagrange’s mean value theorem as it can be seen as a
special case for 𝐹(𝑥) = 𝑥.
Proof
Here 𝐹(𝑏) − 𝐹(𝑎) ≠ 0
Define a new function
𝜙(𝑥) = 𝑓(𝑥) + 𝐴 𝐹(𝑥), 𝑥 ∈ [𝑎, 𝑏]
We choose a number 𝐴 such that the condition 𝜙(𝑎) = 𝜙(𝑏) is satisfied. Then
𝑓(𝑎) + 𝐴𝐹(𝑎) = 𝑓(𝑏) + 𝐴𝐹(𝑏)
𝑓(𝑏) − 𝑓(𝑎)
𝐴=−
𝐹(𝑏) − 𝐹(𝑎)
As a result, we have

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Unit 08: Mean Value Theorems
𝑓(𝑏) − 𝑓(𝑎)
𝜙(𝑥) = 𝑓(𝑥) + (− ) 𝐹(𝑥)
𝐹(𝑏) − 𝐹(𝑎)
The function 𝜙(𝑥) is continuous on the closed interval[𝑎, 𝑏], differentiable on the open interval
(𝑎, 𝑏) and takes equal values at the endpoints of the interval. Therefore, it satisfies all the conditions
of the Rolle’s Theorem. Then there is at least a point 𝑐 in the interval (𝑎, 𝑏) such that
𝜙 ′ (𝑐) = 0
It follows that
𝑓 ′ (𝑐) + 𝐴𝐹 ′ (𝑐) = 0
or
𝑓 ′ (𝑐) = −𝐴𝐹′(𝑐)
or
𝑓 ′ (𝑐) 𝑓(𝑏) − 𝑓(𝑎)
= , 𝐹 ′ (𝑐) ≠ 0
𝐹′(𝑐) 𝐹(𝑏) − 𝐹(𝑎)
Hence the result.

Augustin-Louis Cauchy was one of the greatest mathematicians

during the nineteenth century. In fact, there are sixteen concepts and
theorems named after him, more than any other mathematician. His life
began in Paris, France on August 21, 1789, and ended at Sceaux, France on
May 22, 1857. His father, Luois-Francois, and his mother, Marie-
Madeleine Desestre, provided him and his siblings a comfortable life.
Cauchy was exposed to famous scientists as a child. The Cauchy family
once had Laplace and Berthollet as neighbors, and his father even knew
Lagrange. In fact, Lagrange had foreseen Augustin's scientific greatness
when he was a child by warning his father to not show him any
mathematical text before he was seventeen years old.

8.5 Alternate form of Lagrange’s Mean Value Theorem

We will see the following result without proof.
If two functions 𝑓(𝑥) and 𝐹(𝑥)are continuous on a closed interval [𝑎, 𝑎 + ℎ]and differentiable on the
open interval (𝑎, 𝑎 + ℎ), then there is at least one 𝜃 ∈ (0,1) such that
𝑓(𝑎+ℎ)−𝑓(𝑎) 𝑓′ (𝑎+𝜃ℎ)
= .
𝐹(𝑎+ℎ)−𝐹(𝑎) 𝐹 ′ (𝑎+𝜃ℎ)

If in the Cauchy’s mean value theorem 𝑓(𝑥) = 𝑒 𝑥 and 𝐹(𝑥) = 𝑒 −𝑥 , show that 𝑐 is the
arithmetic mean between 𝑎 and 𝑏.
Here 𝑓 ′ (𝑥) = 𝑒 𝑥
𝐹 ′ (𝑥) = −𝑒 −𝑥
By Cauchy’s mean value theorem,
𝑓 ′ (𝑐) 𝑓(𝑏) − 𝑓(𝑎)
=
𝐹′(𝑐) 𝐹(𝑏) − 𝐹(𝑎)

𝑒𝑐 𝑒𝑏 − 𝑒𝑎
=
−𝑒 −𝑐 𝑒 −𝑏 − 𝑒 −𝑎
−𝑒 2𝑐 = −𝑒 𝑎+𝑏
This implies
𝑎+𝑏
𝑐=
2

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𝑠𝑖𝑛𝛼−𝑠𝑖𝑛𝛽 𝜋
Show that = cot 𝜃, 0<𝛼<𝜃<𝛽<
𝑐𝑜𝑠𝛽−𝑐𝑜𝑠𝛼 2

Looking at the result to be proved, you can see that the left hand side is a ratio of difference of two
functions and Cauchy’s theorem can be used to derive this, provided these two functions satisfy the
requirements of Cauchy’s mean value theorem.
Let 𝑓(𝑥) = sin 𝑥
𝐹(𝑥) = cos 𝑥 𝑥 ∈ [𝛼, 𝛽]
𝑓′(𝑥) = cos 𝑥
𝐹′(𝑥) = −sin 𝑥
By Cauchy’s mean value theorem we can write,
𝑓(𝛽) − 𝑓(𝛼) 𝑓 ′ (𝜃)
=
𝐹(𝛽) − 𝐹(𝛼) 𝐹 ′ (𝜃)
𝑠𝑖𝑛𝛽 − 𝑠𝑖𝑛𝛼 cos 𝜃
=
cos 𝛽 − cos 𝛼 −𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝛽 − 𝑠𝑖𝑛𝛼
= − cot 𝜃
cos 𝛽 − cos 𝛼
or
𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛽
= cot 𝜃
cos 𝛽 − cos 𝛼
Hence the result

Check the validity of Cauchy’s mean value theorem for the functions 𝑓(𝑥) = 𝑥 4 and 𝑔(𝑥) =
𝑥 on the interval[1,2].
2

Here
𝑓(𝑥) = 𝑥 4
𝑓′(𝑥) = 4𝑥 3
𝑔(𝑥) = 𝑥 2
𝑔′(𝑥) = 2𝑥
Both functions are satisfying all the criteria of continuity and differentiability, therefore we can
write
𝑓 ′ (𝑐) 𝑓(𝑏) − 𝑓(𝑎)
=
𝑔′(𝑐) 𝑔(𝑏) − 𝑔(𝑎)
4𝑐 3 𝑏 4 − 𝑎4
= 2
2𝑐 𝑏 − 𝑎2
2𝑐 2 = 𝑎2 + 𝑏 2
Here 𝑎 = 1, 𝑏 = 2

Therefore, 𝑐 = ±√2.5 and √2.5 ∈ (1,2)

Therefore having found at least a value in (1,2) such that
𝑓′ (𝑐) 𝑓(𝑏)−𝑓(𝑎)
= , we can say that the Cauchy’ theorem is valid for the given functions.
𝑔′(𝑐) 𝑔(𝑏)−𝑔(𝑎)

Summary
This unit is an extension of the Rolle’s Theorem. Its generalized form can be seen as Lagrange’s
mean value theorem, which further can be generalized as the Cauchy’s mean value theorem.
• Lagrange’s mean value theorem (MVT) states that if a function 𝑓(𝑥) is continuous on a
closed interval [𝑎, 𝑏]and differentiable on the open interval (𝑎, 𝑏), then there is at least one
𝑓(𝑏)−𝑓(𝑎)
point 𝑥 = 𝑐 on this interval, such that = 𝑓′(𝑐).
𝑏−𝑎

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 Notes
Unit 08: Mean Value Theorems
• If two functions 𝑓(𝑥) and 𝐹(𝑥) are continuous on an interval [𝑎, 𝑏], differentiable on (𝑎, 𝑏)
and F’(x) is non zero for all 𝑥 ∈ (𝑎, 𝑏), then there exists a point 𝑐 ∈ (𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎) 𝑓 ′ (𝑐)
=
𝐹(𝑏) − 𝐹(𝑎) 𝐹 ′ (𝑐)

Key Words
Lagrange’s mean value theorem, Cauchy’s mean value theorem

Self Assessment
on the interval[4, 5].
𝑥−1
1. The Lagrange’s mean value theorem is valid for the function 𝑓(𝑥) =
𝑥−3
A. True
B. False

2. All points c satisfying the conditions of the MVT for the function 𝑓(𝑥) = 𝑥 3 − 𝑥in the interval
[−2, 1] are
A. -1
B. 1, -1
C. 0
D. 1
3. For any quadratic function𝑝𝑥 2 + 𝑞𝑥 + 𝑟, the value of 𝜃 in Lagrange’s theorem, for any
𝑝, 𝑞, 𝑟, 𝑎, ℎ is

A. less than ½
B. greater than ½
C. always ½
D. can take any value

4. For any quadratic function3𝑥 2 + 2𝑥 + 1, the value of 𝜃 in Lagrange’s theorem is

A. 0
B. 1
C. 0.5
D. 1.5

5. Cauchy’s Mean Value Theorem can be reduced to Lagrange’s Mean Value theorem.

A. True
B. False

6. Which of the following is not a necessary condition for Cauchy’s Mean Value Theorem?

A. The functions, f(x) and g(x) be continuous in [a, b]

B. The derivative of g'(x) be equal to 0
C. The functions f(x) and g(x) be derivable in (a, b)
D. There exists a value c ∈ (a, b) such that, (f(b)−f(a)) / (g(b)−g(a)) =f ’ (c) / g ′ (c)

7. Cauchy’s Mean Value Theorem is also known as ‘Extended Mean Value Theorem’.

A. True

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Calculus
B. False

8. The Mean Value Theorem was stated and proved by _______

A. Leonhard Euler
B. Govindasvami
C. Michel Rolle
D. Augustin Louis Cauchy

9. The value of c which satisfies the Mean Value Theorem for the function 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1 on
[1,2] is

A. -5/2
B. -5/2
C. 7/2
D. -7/2

10. What is the value of c which lies in [1, 2] for the function f(x)= 4x and g(x)= 3x 2?
A. 1
B. 1.5
C. 2
D. 2.5

Answer for Self Assessment

1. A 2. A 3. C 4. C 5. A

6. B 7. A 8. D 9. D 10. B

Review Questions
1. State and prove the Lagrange’s mean value theorem.
2. State and prove the Cauchy’s mean value theorem.
3. Check the validity of Cauchy’s MVT for the functions f(x)= 4x and g(x)= 3x2
4. Check the validity of Lagrange’s MVT for the function 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1 on [1,2].
5. Check the validity of Lagrange’s MVT for the function 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1 on [-1,2].
6. Explain how the Lagrange’s MVT is a special case of Cauchy’s MVT.
7. Discuss the Lagrange’s MVT in the interval [𝑎, 𝑎 + ℎ].
8. Discuss the Cauchy’s MVT in the interval [𝑎, 𝑎 + ℎ].

Further Reading
George B. Thomas Jr., Joel Hass, Christopher Heil& Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens& Stephan Davis (2016).Calculus (10th edition).Wiley India.

https://www.math24.net/cauchys-mean-value-theorem
https://www.geeksforgeeks.org/lagranges-mean-value-theorem
https://tutorial.math.lamar.edu/classes/calci/MeanValueTheorem.aspx

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Dr. Pratiksha, Lovely Professional University Unit 09: Higher Order Derivatives

Unit 09: Higher Order Derivatives

CONTENTS
Objectives
Introduction
9.1 Successive derivatives
9.2 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒆𝒂𝒙
9.3 The𝒏𝒕𝒉derivative for 𝒚 = 𝒂𝒙 + 𝒃𝒎 Where 𝒎 is a Positive Integer More than 𝒏
9.4 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒍𝒐𝒈(𝒂𝒙 + 𝒃)
9.5 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒔𝒊𝒏(𝒂𝒙 + 𝒃)
9.6 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒄𝒐𝒔(𝒂𝒙 + 𝒃)
9.7 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒆𝒂𝒙𝒔𝒊𝒏(𝒃𝒙 + 𝒄)
9.8 The 𝒏𝒕𝒉derivative for 𝒚 = 𝒆𝒂𝒙𝒄𝒐𝒔(𝒃𝒙 + 𝒄)
9.9 Determination of 𝒏𝒕𝒉 Derivative of the Rational Functions
9.10 Leibnitz Theorem
Summary
Keywords
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will

• be able to find the 𝑛𝑡ℎ derivative of elementary functions

• be able to find the 𝑛𝑡ℎ derivative of the derived functions
• be able to calculate the 𝑛𝑡ℎ derivative of the product of two functions using the Leibnitz
theorem

Introduction
By now, we have a good idea about what is differentiation, we
know the technique how to differentiate a function, we have
derived quite a few rules of the derivatives for some functions
also. Continuing the stride, we now look into how the higher
derivatives can be found out in a general manner. We will
discuss how to find the nth derivative for some specific
functions in this chapter. We will see the process of
differentiating a given function successively n times, which is
known as successive differentiation and the results that you get
are called successive derivatives.

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The adjacent image gives an idea of the higher derivatives. At first there are coffee beans, then some
changes happen and you see the change in the beans with respect to time (x say), it is the coffee
powder and then some more changes happening to this and you are getting the coffee.
So, we can say that 𝑓′(𝑥) is the first derivative and 𝑓′′(𝑥) is the second derivative. So this second
derivative has come up by differentiating the first derivative. So, this is the successive
differentiation, you can further differentiate to get the third derivative and so, on. This successive
differentiation is very much important for scientific and engineering applications.

9.1 Successive derivatives

The process of differentiating a given function successively n times are called successive
differentiation and the results of such differentiation are called successive derivatives.
Let the function be 𝑦 = 𝑓(𝑥).
𝑑𝑦
Differentiating it once we get = 𝑓 ′ (𝑥)
𝑑𝑥
𝑑𝑦
𝑑( ) 𝑑 2𝑦
Differentiating it twice we get 𝑑𝑥
= = 𝑓′′(𝑥)
𝑑𝑥 𝑑𝑥 2
𝑑2 𝑦
𝑑( ) 𝑑3𝑦
𝑑𝑥2
Differentiating it thrice we get = = 𝑓 ′′ ′(𝑥)
𝑑𝑥 𝑑𝑥 3

and so on
For instance, 𝑓(𝑥) = 𝑥 5 + sin 𝑥 + 𝑒 2𝑥
𝑓 ′ (𝑥) = 5𝑥 4 + cos 𝑥 + 2𝑒 2𝑥
𝑓 ′′ (𝑥) = 20𝑥 3 − sin 𝑥 + 4𝑒 2𝑥
𝑓 ′′′ (𝑥) = 60𝑥 2 − cos 𝑥 + 8 𝑒 2𝑥
and so on.

9.2 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒆𝒂𝒙

We will calculate the 𝑛𝑡ℎ derivative by generalizing first few derivatives.
𝑦1 = 𝑎 𝑒 𝑎𝑥
𝑦2 = 𝑎2 𝑒 𝑎𝑥
𝑦3 = 𝑎3 𝑒 𝑎𝑥
⋮
𝑦𝑛 = 𝑎𝑛 𝑒 𝑎𝑥

9.3 The𝒏𝒕𝒉 derivative for 𝒚 = (𝒂𝒙 + 𝒃)𝒎 Where 𝒎 is a Positive Integer

More than 𝒏
We will calculate the 𝑛𝑡ℎ derivative by generalizing first few derivatives.
𝑦1 = 𝑚𝑎(𝑎𝑥 + 𝑏)𝑚−1
𝑦2 = 𝑚(𝑚 − 1)(𝑎𝑥 + 𝑏)𝑚−2 𝑎2
𝑦3 = 𝑚(𝑚 − 1)(𝑚 − 2)(𝑎𝑥 + 𝑏)𝑚−3 𝑎3
⋮
𝑦𝑛 = 𝑚(𝑚 − 1)(𝑚 − 2) … (𝑚 − 𝑛 + 1)(𝑎𝑥 + 𝑏)𝑚−𝑛 𝑎𝑛
The 𝑛𝑡ℎ derivative can further be written as
𝑚!
𝑦𝑛 = (𝑎𝑥 + 𝑏)𝑚−𝑛 𝑎𝑛
(𝑚 − 𝑛)!

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Unit 09: Higher Order Derivatives
9.4 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒍𝒐𝒈(𝒂𝒙 + 𝒃)
We will calculate the 𝑛𝑡ℎ derivative by generalizing first few derivatives.
𝑦1 = 𝑎(𝑎𝑥 + 𝑏)−1
𝑦2 = −𝑎2 (𝑎𝑥 + 𝑏)−2
𝑦3 = 2! 𝑎3 (𝑎𝑥 + 𝑏)−3
𝑦4 = −3! 𝑎4 (𝑎𝑥 + 𝑏)−4
⋮
𝑦𝑛 = (−1)𝑛−1 𝑎𝑛 (𝑎𝑥 + 𝑏)−𝑛

9.5 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒔𝒊𝒏(𝒂𝒙 + 𝒃)

We will calculate the 𝑛𝑡ℎ derivative by generalizing first few derivatives.
𝜋
𝑦1 = 𝑎 cos(𝑎𝑥 + 𝑏) = 𝑎 sin (𝑎𝑥 + 𝑏 + )
2
𝜋 2𝜋
𝑦2 = 𝑎 2 cos (𝑎𝑥 + 𝑏 + ) = 𝑎2 sin (𝑎𝑥 + 𝑏 + )
2 2
Similarly
3𝜋
𝑦3 = 𝑎3 sin (𝑎𝑥 + 𝑏 + )
2
⋮
𝑛𝜋
𝑦𝑛 = 𝑎𝑛 sin (𝑎𝑥 + 𝑏 + )
2

9.6 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒄𝒐𝒔(𝒂𝒙 + 𝒃)

We will calculate the 𝑛𝑡ℎ derivative by generalizing first few derivatives.
𝜋
𝑦1 = −𝑎 sin(𝑎𝑥 + 𝑏) = 𝑎 cos (𝑎𝑥 + 𝑏 + )
2
𝜋 2𝜋
𝑦2 = −𝑎2 sin (𝑎𝑥 + 𝑏 + ) = 𝑎2 cos (𝑎𝑥 + 𝑏 + )
2 2
Similarly
3𝜋
𝑦3 = 𝑎3 cos (𝑎𝑥 + 𝑏 + )
2
⋮
𝑛𝜋
𝑦𝑛 = 𝑎𝑛 cos (𝑎𝑥 + 𝑏 + )
2

9.7 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒆𝒂𝒙 𝒔𝒊𝒏(𝒃𝒙 + 𝒄)

𝑦1 = 𝑎 𝑒 𝑎𝑥 sin(𝑏𝑥 + 𝑐) + 𝑏 𝑒 𝑎𝑥 cos(𝑏𝑥 + 𝑐)
= 𝑒 𝑎𝑥 (𝑎 sin(𝑏𝑥 + 𝑐) + 𝑏 cos(𝑏𝑥 + 𝑐))
Here substituting 𝑎 = 𝑟 cos 𝜃, 𝑏 = 𝑟 sin 𝜃 in the above expression.
𝑦1 = 𝑒 𝑎𝑥 𝑟 (cos 𝜃 sin(𝑏𝑥 + 𝑐) + sin 𝜃 cos(𝑏𝑥 + 𝑐))
= 𝑟 𝑒 𝑎𝑥 sin(𝑏𝑥 + 𝑐 + 𝜃)
Similarly
𝑦2 = 𝑟 2 𝑒 𝑎𝑥 sin (𝑏𝑥 + 𝑐 + 2𝜃)
⋮
𝑛 𝑎𝑥
𝑦𝑛 = 𝑟 𝑒 sin(𝑏𝑥 + 𝑐 + 𝑛𝜃)

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𝑏
where𝑟 = √𝑎2 + 𝑏 2 and tan 𝜃 =
𝑎
𝑛
𝑏
Therefore, 𝑦𝑛 = (𝑎2 + 𝑏 2 ) eax sin (𝑏𝑥 + 𝑐 + 𝑛𝑡𝑎𝑛−1 ( ))
2
𝑎

9.8 The 𝒏𝒕𝒉 derivative for 𝒚 = 𝒆𝒂𝒙 𝒄𝒐𝒔(𝒃𝒙 + 𝒄)

𝑦1 = −𝑏 𝑒 𝑎𝑥 sin(𝑏𝑥 + 𝑐) + 𝑎 𝑒 𝑎𝑥 cos(𝑏𝑥 + 𝑐)
= 𝑒 𝑎𝑥 (−𝑏 sin(𝑏𝑥 + 𝑐) + 𝑎 cos(𝑏𝑥 + 𝑐))
Here substituting 𝑎 = 𝑟 cos 𝜃, 𝑏 = 𝑟 sin 𝜃 in the above expression.
𝑦1 = 𝑒 𝑎𝑥 𝑟 (−sin 𝜃 sin(𝑏𝑥 + 𝑐) + cos 𝜃 cos(𝑏𝑥 + 𝑐))
= 𝑟 𝑒 𝑎𝑥 cos(𝑏𝑥 + 𝑐 + 𝜃)
Similarly
𝑦2 = 𝑟 2 𝑒 𝑎𝑥 cos (𝑏𝑥 + 𝑐 + 2𝜃)
⋮
𝑦𝑛 = 𝑟 𝑛 𝑒 𝑎𝑥 cos(𝑏𝑥 + 𝑐 + 𝑛𝜃)
𝑏
where𝑟 = √𝑎2 + 𝑏 2 and tan 𝜃 =
𝑎
𝑛
𝑏
Therefore, 𝑦𝑛 = (𝑎2 + 𝑏 2 ) eax cos (𝑏𝑥 + 𝑐 + 𝑛𝑡𝑎𝑛−1 ( ))
2
𝑎

9.9 Determination of 𝒏𝒕𝒉 Derivative of the Rational Functions

To calculate the 𝑛𝑡ℎ derivative of a rational function, we can decompose it into partial fractions. We
may use the De Moivre’s theorem also if the situation demands.

1
Find the 𝑛𝑡ℎ derivative of .
1−5𝑥+6𝑥 2

Now here the given function is a composite function. We can work out on the function to write it as
an elementary function whose 𝑛𝑡ℎ derivative is known.
1
Let 𝑦 =
1−5𝑥+6𝑥 2

1
=
(1 − 3𝑥)(1 − 2𝑥)
3 2
= −
1 − 3𝑥 1 − 2𝑥
1
Now these two expressions are of the form . Working out on the 𝑛𝑡ℎ derivative of this function
𝑎𝑥+𝑏
(−1)𝑛 𝑛!𝑎𝑛
we get (𝑎𝑥+𝑏)𝑛+1

Using this general formula, we can write

𝑛+1 𝑛+1
3 2
𝑦𝑛 = 𝑛! [( ) −( ) ]
1 − 3𝑥 1 − 2𝑥

Find the 𝑛𝑡ℎ derivative of sin 6𝑥 cos 4𝑥

Using the trigonometric identities the given function can be written as y = sin 6𝑥 cos 4𝑥 =
1
(sin 10𝑥 + cos 2𝑥)
2

Applying the direct result of the sine and cosine functions, we get
1 𝑛𝜋 𝑛𝜋
𝑦𝑛 = (10𝑛 sin (10𝑥 + ) + 2𝑛 cos (2𝑥 + ))
2 2 2

𝑑2𝑦
If 𝑦 = 𝑥 + tan 𝑥, show that cos2 𝑥 − 2𝑦 + 2𝑥 = 0
𝑑𝑥 2

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Unit 09: Higher Order Derivatives
We can find the first and second derivatives of 𝑦 and on substituting them in the left hand side, you
can easily get the result.

9.10 Leibnitz Theorem

If u and v are functions of x such that their 𝑛𝑡ℎ derivativesexist, then the 𝑛𝑡ℎ derivative of their
product is given by
(𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶 𝑢𝑛−1 𝑣1 + 𝑛2𝐶 𝑢𝑛−2 𝑣2 + ⋯ + 𝑛𝑟𝐶 𝑢𝑛−𝑟 𝑣𝑟 + ⋯ + 𝑢𝑣𝑛
where𝑢𝑟 and 𝑣𝑟 represent the 𝑟 𝑡ℎ derivatives of 𝑢 and 𝑣 respectively.
𝑑 𝑑𝑣 𝑑𝑢
(𝑢𝑣) = 𝑢 +𝑣
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑2 𝑑
(𝑢𝑣) = (𝑢𝑣1 + 𝑣𝑢1 ) = 𝑢2 𝑣 + 2𝑢1 𝑣1 + 𝑢𝑣2
𝑑𝑥 2 𝑑𝑥
and continuing in the same manner, the 𝑛𝑡ℎ derivative can be obtained as
𝑑𝑛
(𝑢𝑣) = (𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶 𝑢𝑛−1 𝑣1 + 𝑛2𝐶 𝑢𝑛−2 𝑣2 + ⋯ + 𝑛𝑟𝐶 𝑢𝑛−𝑟 𝑣𝑟 + ⋯ + 𝑢𝑣𝑛
𝑑𝑥 𝑛
Let us see some examples to understand the theorem better!

Find the 𝑛𝑡ℎ derivative of 𝑥 log 𝑥.

Here we can see the given function as a product of two functions and Leibnitz theorem can be
applied.
Let 𝑢 = log 𝑥
1
Then 𝑢1 =
𝑥

1
𝑢2 = −
𝑥2
2
𝑢3 = 3
𝑥
2.3
𝑢4 = − 4
𝑥
⋮
(−1)𝑛−1 (𝑛 − 1)!
𝑢𝑛 =
𝑥𝑛
Let 𝑣 = 𝑥
𝑣1 = 1
𝑣2 = 0
𝑣3 = 0
⋮
𝑣𝑛 = 0
By Leibnitz theorem,
(𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶 𝑢𝑛−1 𝑣1 + 𝑛2𝐶 𝑢𝑛−2 𝑣2 + ⋯ + 𝑛𝑟𝐶 𝑢𝑛−𝑟 𝑣𝑟 + ⋯ + 𝑢𝑣𝑛
(−1)𝑛−1 (𝑛 − 1)! 𝑛(−1)𝑛−2 (𝑛 − 2)!
(𝑥𝑙𝑜𝑔𝑥)𝑛 = +
𝑥 𝑛−1 𝑥 𝑛−1
(−1)𝑛−2 (𝑛 − 2)!
=
𝑥 𝑛−1

Find the 𝑛𝑡ℎ derivative of 𝑥 2 𝑒 3𝑥 𝑠𝑖𝑛4𝑥

Let 𝑢 = 𝑒 3𝑥 𝑠𝑖𝑛4𝑥 and 𝑣 = 𝑥 2
By Leibnitz theorem,

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Calculus
(𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶 𝑢𝑛−1 𝑣1 + 𝑛2𝐶 𝑢𝑛−2 𝑣2 + ⋯ + 𝑛𝑟𝐶 𝑢𝑛−𝑟 𝑣𝑟 + ⋯ + 𝑢𝑣𝑛
4
(𝑥 2 𝑒 3𝑥 𝑠𝑖𝑛4𝑥)𝑛 = 𝑒 3𝑥 5𝑛 sin (4𝑥 + 𝑛𝑡𝑎𝑛−1 ( )) 𝑥 2 + 𝑛 𝑒 3𝑥 5𝑛−1 sin (4𝑥
3
−1
4 𝑛(𝑛 − 1) 3𝑥 𝑛−2 4
+ (𝑛 − 1) tan ) (2𝑥) + 𝑒 5 sin (4𝑥 + (𝑛 − 2) tan−1 ) (2)
3 2 3
The right hand side expression can be simplified for a concise form.

If 𝑦 = tan−1 𝑥, show that (1 + 𝑥 2 )𝑦𝑛+2 + 2(𝑛 + 1)𝑥𝑦𝑛+1 + 𝑛(𝑛 + 1)𝑦𝑛 = 0. Also find 𝑦𝑛 (0).
Here 𝑦 = tan−1 𝑥
1
𝑦1 =
1 + 𝑥2

⇒ (1 + 𝑥 2 )𝑦1 = 1

Differentiating both sides w.r.t. x, we get

(1 + 𝑥 2 )𝑦2 + 2𝑥𝑦1 = 0
Differentiating ‘n’ times w.r.t. x, we get
(1 + 𝑥 2 )𝑦𝑛+2 + 2𝑥 𝑛 𝑦𝑛+1 + 𝑛(𝑛 − 1)𝑦𝑛 + 2(𝑥 𝑦𝑛+1 + 𝑛 𝑦𝑛 ) = 0
(1 + 𝑥 2 )𝑦𝑛+2 + 2𝑥(𝑛 + 1)𝑦𝑛+1 + 𝑛(𝑛 + 1)𝑦𝑛 = 0
which is the required expression to be proved. Now in order to deduce the second part, let us put
𝑥 = 0 in the expressions of 𝑦, 𝑦1 , 𝑦2 and 𝑦𝑛+2 , we get
𝑦(0) = 0
𝑦1 (0) = 1
𝑦2 (0) = 0
𝑦𝑛+2 (0) = −𝑛(𝑛 + 1)𝑦𝑛 (0)
From this recursion formula, higher derivatives can be obtained.
𝑦3 (0) = −1.2. 𝑦1 (0) = −2.1 = −2!
𝑦4 (0) = −2.3. 𝑦2 (0) = 0
𝑦5 (0) = −3.4. 𝑦3 (0) = −3.4. (−2) = 4!
𝑦6 (0) = −4.5. 𝑦4 (0) = 0
𝑦7 (0) = −5.6. 𝑦5 (0) = −5.6.4! = −6!
⋮
𝑦2𝑛+1 (0) = (−1)𝑛 (2𝑛)! and 𝑦2𝑛 (0) = 0
This expression shows that all the even derivatives of the given function are zero at 𝑥 = 0 and the
odd derivatives are given by 𝑦2𝑛+1 (0) = (−1)𝑛 (2𝑛)!

Summary
This chapter is about the higher derivative of a function. We also learnt about how to find the nth
derivative of the product of two functions.

• The process of differentiating a given function successively n times are called successive
differentiation and the results of such differentiation are called successive derivatives.
• If u and v are functions of x such that their 𝑛𝑡ℎ derivatives exist, then the 𝑛𝑡ℎ derivative of
their product is given by
(𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶 𝑢𝑛−1 𝑣1 + 𝑛2𝐶 𝑢𝑛−2 𝑣2 + ⋯ + 𝑛𝑟𝐶 𝑢𝑛−𝑟 𝑣𝑟 + ⋯ + 𝑢𝑣𝑛
where𝑢𝑟 and 𝑣𝑟 represent the 𝑟 𝑡ℎ derivatives of 𝑢 and 𝑣 respectively.

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Unit 09: Higher Order Derivatives
Keywords
Successive differentiation, higher derivatives, Leibnitz theorem

Self Assessment
1. The derivative of the function √2 + 𝑥 2 is
𝑥
A. 2
2√2+𝑥
𝑥
B.
√2+2𝑥 2
2𝑥
C.
√2+𝑥 2
𝑥
D.
√2+𝑥 2

𝑑
2. (𝑠𝑖𝑛(𝑐𝑜𝑠𝑥)) =
𝑑𝑥
A. sin 𝑥 cos (cos 𝑥)
B. cos (sin 𝑥)
C. −sin 𝑥 cos (cos 𝑥)
D. sin 𝑥 cos (sin 𝑥)

3. If f is differentiable on the closed interval [a, b] and r is any number between f ’ (a) and
f ’ (b), then there exists a number c in the open interval (a, b) such that f ’(c) = r. This
statement is of
A. intermediate value theorem
B. mean value theorem
C. Rolle’s theorem
D. Darboux’s theorem

−1 𝑥
𝑒 𝑠𝑖𝑛ℎ
4. is the derivative of the function
√1+𝑥 2
−1
𝑒 𝑠𝑖𝑛ℎ 𝑥
A. x
√1+𝑥 2
−1 𝑥
𝑒 𝑠𝑖𝑛ℎ
B.
√1+𝑥 2
𝑠𝑖𝑛ℎ−1 𝑥
C. 𝑒 ans
−1 𝑥
D. 𝑒 𝑐𝑜𝑠ℎ

5. The derivative of 𝑙𝑜𝑔(𝑐𝑜𝑠ℎ 𝑥) w.r.t. 𝑥 is

A. log (sin 𝑥)
B. log (sinh 𝑥)
C. 𝑐𝑜𝑠𝑒𝑐ℎ 𝑥
D. tanh 𝑥

𝑑 𝑠𝑖𝑛𝑥
6. 𝑒 =
𝑑𝑥
A. 𝑒 𝑠𝑖𝑛𝑥 cos 𝑥
B. cos (sin 𝑥)
C. − 𝑒 𝑠𝑖𝑛𝑥 cos 𝑥
D. sin 𝑥 cos (sin 𝑥)

7. The Leibnitz theorem is about

A. the 𝑛𝑡ℎ derivative of the sum of two functions
B. the 𝑛𝑡ℎ derivative of the difference of two functions
C. the 𝑛𝑡ℎ derivative of the quotient of two functions
D. the 𝑛𝑡ℎ derivative of the product of two functions

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8. Which of the following is correct?
A. (𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶𝑢𝑛−1 𝑣1 + 𝑛2𝐶𝑢𝑛−2 𝑣2 + 𝑛3𝐶𝑢𝑛−3 𝑣3 + … + 𝑛𝑟𝐶𝑢𝑛−𝑟 𝑣𝑟 + … + 𝑢 𝑣𝑛
B. (𝑢𝑣)𝑛 = 𝑢 𝑣 + 𝑛1𝐶𝑢𝑛−1 𝑣1 + 𝑛2𝐶𝑢𝑛−2 𝑣2 + 𝑛3𝐶𝑢𝑛−3 𝑣3 + … + 𝑛𝑟𝐶𝑢𝑛−𝑟 𝑣𝑟 + … + 𝑢 𝑣𝑛
C. (𝑢𝑣)𝑛 = 𝑢𝑛 𝑣 + 𝑛1𝐶𝑢𝑛−1 𝑣1 + 𝑛2𝐶𝑢𝑛−2 𝑣2 + 𝑛3𝐶𝑢𝑛−3 𝑣3 + … + 𝑛𝑟𝐶𝑢𝑛−𝑟 𝑣𝑟 + … + 𝑢 𝑣
D. (𝑢𝑣)𝑛 = 𝑢 𝑣 + 𝑛1𝐶𝑢𝑛−1 𝑣1 + 𝑛2𝐶𝑢𝑛−2 𝑣2 + 𝑛3𝐶𝑢𝑛−3 𝑣3 + … + 𝑛𝑟𝐶𝑢𝑛−𝑟 𝑣𝑟 + … + 𝑣 𝑢

𝒔𝒊𝒏𝒙
9. Let 𝒇(𝒙) = . The first derivative of 𝒇(𝒙) at 𝒙 = 𝟎 is given by
𝟏+𝒙𝟐

A. 1
B. 0
C. -1
D. 2

10. The number of terms in the 𝑛𝑡ℎ derivative of 𝑥 2 𝑒 3𝑥 𝑠𝑖𝑛 4𝑥 are

A. 1
B. 2
C. 3
D. 4

11. For 𝑦 = 𝑡𝑎𝑛−1 𝑥, (1 + 𝑥 2 )𝑦𝑛+2 + 2(𝑛 + 1)𝑥𝑦𝑛+1 + 𝑛(𝑛 + 1)𝑦𝑛 = 0, Then 𝑦3 (0) is
A. 0
B. 1
C. 2
D. -2

𝑑
12. 𝑒 −𝑠𝑖𝑛𝑥 =
𝑑𝑥
A. −𝑒 −𝑠𝑖𝑛𝑥 cos 𝑥
B. cos (sin 𝑥)
C. − 𝑒 𝑠𝑖𝑛𝑥 cos 𝑥
D. sin 𝑥 cos (sin 𝑥)

Answer for Self Assessment

1. D 2. C 3. D 4. C 5. D

6. A 7. D 8. A 9. A 10. C

11. D 12. A

Review Questions
1. Find the first three derivatives of the following expressions w.r.t. x
𝑥 2 +𝑎
(i)
𝑥+𝑎
2
(ii) 8𝑥 + 3.8𝑥 3 − 𝑥 2 + 𝑥 − 7
4
3

2. If a body move according to the law

𝑠 = 12 − 4.5𝑡 + 6.2𝑡 2
find its velocity and acceleration when t=4 seconds, 𝑠 being in feet. Is the acceleration the
same for all values of 𝑡?

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Unit 09: Higher Order Derivatives
3. Find the 𝒏𝒕𝒉 derivative for 𝒚 = 𝒆𝒂𝒙 cos(𝒃𝒙 + 𝒄)
4. Find the 𝒏𝒕𝒉 derivative for 𝒚 = 𝒆𝒂𝒙 sin(𝒃𝒙 − 𝒄)
1+𝑥
5. Find the 𝒏𝒕𝒉 derivative for 𝒚 = tan−1
1−𝑥
1
6. Find the 𝒏𝒕𝒉 derivative for 𝒚 =
a2 −𝑥 2

7. State and prove Leibnitz theorem.

𝑑𝑛
8. If 𝐼𝑛 = (𝑥 𝑛 log 𝑥) , prove that 𝐼𝑛 = 𝑛𝐼𝑛−1 + (𝑛 − 1)
𝑑𝑥 𝑛
−1 𝑥
9. Find the value of 𝒏𝒕𝒉 derivative for 𝒚 = emsin for 𝑥 = 0.
−1
10. If 𝒚 = easin 𝑥 prove that
(1 − 𝑥 2 )𝑦𝑛+2 − (2𝑛 + 1)𝑥𝑦𝑛+1 − (𝑛2 + 𝑎2 )𝑦𝑛 = 0

Further Reading
George B. Thomas Jr., Joel Hass, Christopher Heil& Maurice D. Weir (2018). Thomas’
Calculus (14th edition). Pearson Education.
Howard Anton, I. Bivens& Stephan Davis (2016).Calculus (10th edition).Wiley India.

http://www.calculusmadeeasy.org/7.html
https://www.math24.net/leibniz-formula

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Dr. Pratiksha, Lovely Professional University Unit 10: Maclaurin’s and Taylor’s Theorems

Unit 10: Maclaurin’s and Taylor’s Theorems

CONTENTS
Objectives
Introduction
10.1 Generalized Mean Value Theorem- Taylor’s Theorem
10.2 Maclaurin’s Theorem
10.3 Taylor’s Theorem in Finite form with Lagrange form of Remainder
10.4 Maclaurin's Power Series for a Given Function
10.5 Taylor’s Theorem in Finite form with Cauchy forms of Remainder
Summary
Keywords
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will be able to
 expand the functions using Maclaurin’s theorem
 expand the functions using Taylor’s theorem
 apply the Taylor’s theorem in finite form with Lagrange form of remainder
 apply the Taylor’s theorem in finite form with Cauchy form of remainder

Introduction
In calculus, Taylor's theorem gives us a polynomial which approximates the function in terms of
the derivatives of the function. Since the derivatives are usually easy to compute, these polynomials
are also easy to compute.
A simple example of Taylor's theorem is the approximation of the exponential function 𝑒 near
𝑥 = 0. In other words, the exponential function can be approximated by an infinite polynomial
given as follows
𝑥 𝑥 x
𝑒 = 1+𝑥+ + + ⋯+ +⋯
2! 3! 𝑛!
For a derivable function𝑓, we can say that 𝑓′ exists in certain neighborhood of point 𝑐 and this
further implies that 𝑓 is defined and is continuous in a neighborhood of 𝑐.
Similarly, if 𝑓′ has derivative at 𝑐 has the same meaning as 𝑓 has a second derivative at 𝑐. And this
further implies that 𝑓′ is continuous at 𝑐.
In general if 𝑓 (𝑥) exists in the neighborhood of 𝑐, then the derivative of 𝑓 (𝑥) at 𝑐, if exists, is
called the 𝑛 derivative of 𝑓 at 𝑐 and is written as 𝑓 ( ) (𝑐).

10.1 Generalized Mean Value Theorem- Taylor’s Theorem

If 𝑛 ≥ 0 is an integer and 𝑓 is a function which is 𝑛 times continuously differentiable on the closed
interval[𝑎, 𝑥] and 𝑛 + 1 times differentiable on the open interval(𝑎, 𝑥), then

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𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎)
𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) + (𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎) + 𝑅 (𝑥)
1! 2! 𝑛!
Here, 𝑛! denotes the factorial of , and 𝑅 (𝑥) is a remainder term, denoting the difference between
the Taylor polynomial of degree 𝑛 and the original function. The remainder term 𝑅 (𝑥)depends on
𝑥and is small if 𝑥is close enough to 𝑎. There are several expressions available for it.
We can state the theorem in the following form also.
If a function 𝑓 is such that
(i) the (𝑛 − 1) derivative 𝑓 is continuous in [𝑎, 𝑎 + ℎ],
(ii) the 𝑛 derivative 𝑓 exists in (𝑎, 𝑎 + ℎ) and
(iii) 𝑝 is a given positive integer
Then there exists at least one 𝜃 ∈ (0,1) such that
( ) ( ) ( ) ( )
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ !
+ℎ !
+⋯+ℎ !
+ ( )!
𝑓 (𝑎 + 𝜃ℎ) ---- (1)

Condition (i) assures that 𝑓, 𝑓 , 𝑓 , … , 𝑓 are continuous in [𝑎, 𝑎 + ℎ].

Let a function 𝜙(𝑥) be defined by
(𝑎 + ℎ − 𝑥) (𝑎 + ℎ − 𝑥)
𝜙(𝑥) = 𝑓(𝑥) + (𝑎 + ℎ − 𝑥)𝑓 (𝑥) + 𝑓 (𝑥) + ⋯ + 𝑓 (𝑥) + 𝐴(𝑎 + ℎ − 𝑥)
2! (𝑛 − 1)!
---- (2)
where𝐴 is to be determined such that
𝜙(𝑎) = 𝜙(𝑎 + ℎ)
Therefore,
𝜙(𝑥)is a continuous in [𝑎, 𝑎 + ℎ], derivable in (𝑎, 𝑎 + ℎ) and 𝜙(𝑎) = 𝜙(𝑎 + ℎ). So, Rolle’s theorem
suggests that there must exist at least a 𝜃 ∈ (0,1) such that
𝜙 (𝑎 + 𝜃ℎ) = 0
( )
Now 𝜙 (𝑥) = 𝑓 (𝑥) + (𝑎 + ℎ − 𝑥)𝑓 (𝑥) − 𝑓 (𝑥) + 𝑓 (𝑥) − (𝑎 + ℎ − 𝑥)𝑓 (𝑥) + ⋯ +
!
( ) ( )( )
( )!
𝑓 (𝑥) − ( )!
𝑓 (𝑥) − 𝑝𝐴(𝑎 + ℎ − 𝑥)

(𝑎 + ℎ − 𝑥)
𝜙 (𝑥) = 0 ⇒ 𝑓 (𝑥) = 𝑝𝐴(𝑎 + ℎ − 𝑥)
(𝑛 − 1)!
(𝑎 + ℎ − 𝑎 − 𝜃ℎ)
𝜙 (𝑎 + 𝜃ℎ) = 0 ⇒ 𝑓 (𝑎 + 𝜃ℎ) = 𝑝𝐴(𝑎 + ℎ − 𝑎 − 𝜃ℎ)
(𝑛 − 1)!
(ℎ − 𝜃ℎ)
⇒ 𝑓 (𝑎 + 𝜃ℎ) = 𝑝𝐴(ℎ − 𝜃ℎ)
(𝑛 − 1)!
(1 − 𝜃)ℎ
⇒ 𝐴= 𝑓 (𝑎 + 𝜃ℎ), 1 − 𝜃 ≠ 0, ℎ≠0
(𝑛 − 1)! 𝑝
Substituting 𝐴 in the expression (2), we get the required result.
Corollary
Let 𝑥 be a point of the interval [𝑎, 𝑎 + ℎ]. Let 𝑓 satisfies the conditions of Taylor’s theorem in
[𝑎, 𝑎 + ℎ], thus it satisfies the condition for [𝑎, 𝑥] also.
Writing 𝑎 + ℎ as 𝑥 or ℎ as 𝑥 − 𝑎 in the expression (1), we get
𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎) (𝑥 − 𝑎) (1 − 𝜃)
𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) + (𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎) + 𝑓 (𝑎 + 𝜃(𝑥
1! 2! 𝑛! (𝑛 − 1)! 𝑝
− 𝑎))
---- (3)
where0 < 𝜃 < 1 and the expression (3) holds for all 𝑥 ∈ [𝑎, 𝑎 + ℎ].

10.2 Maclaurin’s Theorem

Substituting 𝑎 = 0 in (3) i.e. for all 𝑥 ∈ [0, ℎ]

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Unit 10: Maclaurin’s and Taylor’s Theorems
( ) ( ) ( ) ( ) ( )
𝑓(𝑥) = 𝑓(0) + (𝑥) + (𝑥) + ⋯ + (𝑥)
( ) + ( )!
𝑓 (𝜃𝑥) ---- (4)
! ! ( )!

which holds when

(i) 𝑓 is continuous in [0, ℎ]]
(ii) 𝑓 exists in (0, ℎ) and
(iii) 𝑝is a given positive integer.

Show that
𝑥 𝑥 x 𝑥
𝑒 =1+𝑥+ + + ⋯+ + 𝑒
2! 3! (𝑛 − 1)! 𝑛!
Here𝑓(𝑥) = 𝑒
𝑓 (𝑥)is continuous in [0, ℎ]
𝑓 (𝑥)exists in (0, ℎ)
Let 𝑝 = 𝑛 in (4). Then,
𝑓 (𝑥)
(𝑥 = 𝑒 𝑓 (0) = 1
𝑓 (𝑥)
(𝑥 = 𝑒 𝑓 ′(0) = 1
𝑓 (𝑥)
(𝑥 = 𝑒 𝑓 (0) = 1
𝑓 (𝑥)
(𝑥 = 𝑒 𝑓 (0) = 1
𝑓 (𝑥)
( ) = 𝑒 𝑓 (𝜃𝑥) = 𝜃𝑥
Therefore, fromthe expression (4), we get
𝑥 𝑥 x 𝑥
𝑒 =1+𝑥+ + + ⋯+ + 𝑒
2! 3! (𝑛 − 1)! 𝑛!
Hence the proof.

10.3 Taylor’s Theorem in F

Finite form with Lagrange form of
Remainder
From the previous section we know, Taylor’s theorem states that, if
i a function 𝑓 is such that
(i) the (𝑛 − 1) derivative 𝑓 is continuous in [𝑎, 𝑎 + ℎ],
(ii) the 𝑛 derivative 𝑓 exists in (𝑎,( 𝑎 + ℎ) and
(iii) 𝑝 is a given positive integer
Then there exists at least one 𝜃 ∈ (0,1) such that
𝑓 (𝑎) 𝑓 (𝑎)
( 𝑓 (𝑎) ℎ (1 − 𝜃)
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ + ⋯+ ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2
2! 𝑛! (𝑛 − 1)! 𝑝
( )
The term ( )!
𝑓 (𝑎 + 𝜃ℎ) is known as the
th remainder after n terms, better known as Taylor’s
remainder 𝑅 after 𝑛 terms due to Schlomilch and Roche. In this expression if we substitute 𝑝 = 𝑛

𝑅 = 𝑓 (𝑎 + 𝜃ℎ)is the remainder after 𝑛 terms due to Lagrange.

!

Therefore the Taylor’s theorem

m with Lagrange’s form of remainder is given as,
𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎) ℎ
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ + ⋯+ ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2! (𝑛 − 1)! 𝑛!
or
( ) ( ) ( ) ( )
𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) + (𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎) ( )!
+ 𝑓 (𝑎 + 𝜃(𝑥 − 𝑎))
! ! !
----- (5)

The expression of 𝑓(𝑥) in (5) will fail for those values of 𝑥 for which
(i) or any of its differential coefficient becomes infinite.
𝑓(𝑥)or
(ii) 𝑓(𝑥) or any of its differential coefficients is discontinuous and

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(iii) lim → 𝑅 ≠ 0 i.e. lim !
𝑓 (𝑎 + 𝜃ℎ) ≠ 0
→

Similarly the expansion of 𝑓(𝑥) by Maclaurin’s theorem is not valid for the values of 𝑥 for which
(i) 𝑓(0) or any of 𝑓 (0), 𝑓 (0), … is not finite
(ii) 𝑓(𝑥) or any of its derivatives is discontinuous as 𝑥 passes through zero and
(iii) lim → 𝑅 ≠ 0 i.e. lim 𝑓 (𝜃𝑥) ≠ 0
→ !

10.4 Maclaurin's Power Series for a Given F

Function
Let a function 𝑓 possesses continuous derivatives of all orders in the interval [0
0, 𝑥], so that we have
𝑥
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 (0) + ⋯ + 𝑓 (0) + 𝑅
(𝑛 − 1)!
where𝑅 is the Lagrange form of remainder.

Therefore 𝑓(𝑥) = 𝑓(0)

( ) + 𝑥𝑓 (0) + ⋯ +
!
𝑓 (0) + ⋯ ---- (6)

is valid for all values of 𝑛 for which lim → 𝑅 = 0.. The expression (6) is called Maclaurin’s infinite
series for the expansion of 𝑓(𝑥) as power series.

Consider the function 𝑓(𝑥) = 𝑒

𝑓 (𝑥) = 𝑒 ∀ 𝑥 ∈ 𝑹
The Lagrange’s form of remainder after 𝑛 terms is
𝑥 𝑥
𝑅 = 𝑓 (𝜃𝑥) = 𝑒 where 0 < 𝜃 < 1
𝑛! 𝑛!
Now consider the case if 𝑥 > 0
𝜃𝑥 < 𝑥 ⇒ 𝑒 <𝑒

And if 𝑥 < 0

Then – 𝑥 > 0
Therefore 𝜃 > 0
⇒ −𝜃𝑥 > 0

⇒𝑒 >𝑒

⇒𝑒 <1

Assuming that for all 𝑥

lim → =0 (The proof is given after the next example)

!

Therefore 𝑅 → 0 as 𝑛 → ∞ ∀ 𝑥 ∈ 𝑹

∴𝑒 =1+𝑥+ +⋯+ + ⋯is valid for all 𝑥 ∈ 𝑹

! !

Consider the function 𝑓(𝑥) = sin 𝑥

𝑛𝜋
𝑓 (𝑥) = sin 𝑥 + ∀𝑥∈𝑹
2
Lagrange’s form of remainder
𝑥 𝑥 𝑛𝜋
𝑅 = 𝑓 (𝜃𝑥) = sin 𝜃𝑥 +
𝑛! 𝑛! 2
𝑥 𝑛𝜋 𝑥
|𝑅 | = sin 𝜃𝑥 + ≤
𝑛! 2 𝑛!
and𝑅 → 0 as 𝑛 → ∞ ∀ 𝑥 ∈ 𝑹
Thus

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Unit 10: Maclaurin’s and Taylor’s Theorems
∀ 𝑥∈𝑹
𝑥 𝑥
𝑠𝑖𝑛𝑥 = 𝑥 − + −⋯
3! 5!

We can prove lim → as follows:

= 0 ∀ 𝑥 ∈ 𝑹as
!

Let 𝑎 = ∀ 𝑥 ∈ 𝑹 𝑎𝑛𝑑 𝑛 ∈ 𝑵
!

If 𝑥 = 0 lim → 𝑎 =0
If 𝑥 > 0 then for ∈ 𝑵 , 𝑎 > 0
For sufficiently large 𝑛 (say 𝑛 ≥ 𝑥)
𝑥 𝑥
𝑎 = = 𝑎 <𝑎
((𝑛 + 1)! 𝑛 + 1
This implies that after certain 𝑛, 𝑎 <𝑎
Since a bounded monotonically decreasing sequence of real numbers must have a limit,
𝑎 = lim 𝑎 = lim 𝑎
→ →
𝑥
= lim lim 𝑎
→ 𝑛+1 →

⇒ 𝑎=0

If 𝑥 < 0, we introduce 𝑎(−1) factor

( )
i.e. = where (−1) is bounded and tends to zero.
! ! !

{𝑏 }is bounded and 𝑎 → 0, then lim → 𝑎 𝑏 =0

Therefore lim → =0∀ 𝑥∈𝑹

!

Hence proved

10.5 Taylor’s Theorem in Finite

F form with Cauchy forms of
Remainder
The Taylor’s theorem states that, if a function 𝑓 is such that
(i) the (𝑛 − 1) derivative 𝑓 is continuous in [𝑎, 𝑎 + ℎ],
(ii) the 𝑛 derivative 𝑓 exists in (𝑎,( 𝑎 + ℎ) and
(iii) 𝑝 is a given positive integer
Then there exists at least one 𝜃 ∈ (0,1) such that
𝑓 (𝑎) 𝑓 (𝑎)
( 𝑓 (𝑎) ℎ (1 − 𝜃)
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ + ⋯+ ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2
2! 𝑛! (𝑛 − 1)! 𝑝
( )
The term ( )!
𝑓 (𝑎 + 𝜃ℎ) is known as the remainder after n terms, better known as Taylor’s
remainder 𝑅 after 𝑛 terms due to Schlomilch and Roche. In this expression if we substitute 𝑝 = 1
( )
𝑅 = ( )!!
is the remainder after 𝑛 terms due to Cauchy.
𝑓 (𝑎 + 𝜃ℎ)is

Therefore the
he Taylor’s theorem with Cauchy
Cauchy’s form of remainder is given as,
𝑓 (𝑎) 𝑓 (𝑎)
( ) 𝑓 (𝑎) ℎ (1 − 𝜃)
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ +⋯+ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2! (𝑛 − 1)! (𝑛 − 1)!!
or
( ) ( ) ( ) ( ) ( )
𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) + (𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎) ( )!
+ ( )!!
𝑓 (𝑎 + 𝜃(𝑥 −
! !
𝑎)) ----- (6)

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Calculus
Expansion of (𝟏 + 𝒙)𝒎 , 𝒎 ∈ 𝑹
(1 + 𝑥) possesses continuous derivatives of every order when 1 + 𝑥 > 0 i.e. 𝑥 > −1. Also
𝑓 (𝑥) = 𝑚(𝑚 − 1)(𝑚 − 2) … (𝑚 − 𝑛 + 1)(1 + 𝑥)
𝑥
𝑅 = (1 − 𝜃) 𝑓 (𝜃𝑥)
(𝑛 − 1)!
𝑥
= (1 − 𝜃) 𝑚(𝑚 − 1) … (𝑚 − 𝑛 + 1)(1 + 𝜃𝑥)
(𝑛 − 1)!
𝑚(𝑚 − 1) … (𝑚 − 𝑛 + 1) 1 − 𝜃
=𝑥 (1 + 𝜃𝑥)
(𝑛 − 1)! 1 + 𝜃𝑥
Let |𝑥| < 1
⇒−1 <𝑥 < 1

Now −1 < 𝑥
⇒ −𝜃 < 𝜃𝑥

⇒ 1 − 𝜃 < 1 + 𝜃𝑥

1−𝜃
⇒ <1
1 + 𝜃𝑥
1−𝜃
⇒0< <1
1 + 𝜃𝑥
Let 𝑚 − 1 > 0, we have
0<𝜃<1
⇒ 𝜃𝑥 < 𝑥

⇒ 𝜃𝑥 + 1 < 𝑥 + 1

Moreover 𝑥 > −1 ⇒ 𝑥 < 1

⇒ 𝜃𝑥 + 1 < 2

Therefore
0 < 𝜃𝑥 + 1 < 2
⇒ 0 < (𝜃𝑥 + 1) <2 <2

Let 𝑚 − 1 < 0, we have

𝜃𝑥 > −|𝑥|
⇒ 𝜃𝑥 + 1 > 1 − |𝑥|

⇒ (𝜃𝑥 + 1) ≤ (1 − |𝑥|)

( )…( )
We know, lim → ( )!
=0

∴ 𝑅 → 0 as 𝑛 → ∞ if|𝑥| < 1
( )
∴ (1 + 𝑥) = 1 + 𝑚𝑥 + !
𝑥 + ⋯when−1 < 𝑥 < 1

Expansion of 𝒍𝒐𝒈(𝟏 + 𝒙)
(1 + 𝑥) possesses continuous derivatives of every order when 1 + 𝑥 > 0 i.e. 𝑥 > −1. Also
(−1) (𝑛 − 1)!
𝑓 (𝑥) =
(1 + 𝑥)
Taking Cauchy’s form of remainder

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Unit 10: Maclaurin’s and Taylor’s Theorems
𝑥
𝑅 = (1 − 𝜃) 𝑓 (𝜃𝑥)
(𝑛 − 1)!
𝑥 (−1) (𝑛 − 1)!
= (1 − 𝜃)
(𝑛 − 1)! (1 + 𝜃𝑥)
1 1−𝜃
= (−1) 𝑥
1 + 𝜃𝑥 1 + 𝜃𝑥
Let |𝑥| < 1
⇒−1<𝑥 <1

⇒−θ< 𝜃𝑥 <𝜃

⇒1 −θ < 1+ 𝜃 𝑥 < 1 + 𝜃

1−θ
⇒0 < <1
1 + θx
1−θ
⇒ <1
1 + θx
Also we have
𝜃𝑥 > −|𝑥|
⇒ 1 + 𝜃𝑥 > 1 − |𝑥|

1 1
⇒ <
1 + 𝜃𝑥 1 − |𝑥|
Therefore, for all 𝑛
1
|𝑅 | < |𝑥| → 0 as 𝑛 → ∞
1 − |𝑥|
Therefore, when |𝑥| < 1,

𝑥 𝑥 (−1) 𝑥
log(1 + 𝑥) = 𝑥 − + − ⋯+ +⋯
2 3 𝑛−1

By taking the Lagrange’s form of remainder we may show that the infinite series expansion is valid
for 𝑥 = 1 also.
For the formal expansion of a function, we will follow the following steps:

Calculate the 𝑛 derivative of the function

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Calculus

Check the lim → 𝑅

If lim → 𝑅 vanishes, then the function can be

expressed as a power series

If 𝑓(𝑥) can be expressed as an infinite Maclaurin’s series, then

𝑥
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 (0) + 𝑓 (0) + ⋯
2!

Can 𝑓(𝑥) = 𝑒 𝑥 ≠ 0 be expanded by Maclaurin’s theorem?

0 𝑥=0
In this problem we need to check all the conditions first regarding the function’s continuity anf
differentiability on its domain.
At the point 𝑥 = 0, we can observe the limit of the function as
1 1 1
lim 𝑓(𝑥) = lim 𝑒 = lim (1 + + + +⋯) = ∞
→ → → 𝑥 2𝑥 6𝑥
As the limit of the given function
functi is not defined at 𝑥 = 0,, the continuity of the function can not be
established, so the given function can not be expanded by Maclaurin’s theorem

Can 𝑓(𝑥) = √𝑥 be expanded by Maclaurin’s theorem?

Clearly the function is a continuous one on its domain.
domain. Let us check the differentiability also.
We have
1 1
𝑓 (𝑥) = 𝑥 =
2 2 √𝑥
1 1
𝑓 (𝑥) = − 𝑥 =−
4 4𝑥 √𝑥
⋮
Clearly 𝑓 (0), 𝑓 ((0),
) … do not exist.

Use Maclaurin’s theorem to expand 𝑦 = log(1 + 𝑒 ) ------ (1)

⇒𝑒 = 1+𝑒

Differentiating both sides w.r.t.𝑥

w.r.t.
𝑒 𝑦 =𝑒 ---- (2)
Differentiating again both sides w.r.t.
w.r.t.𝑥

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Unit 10: Maclaurin’s and Taylor’s Theorems
𝑒 𝑦 +𝑒 𝑦 =𝑒 ---- (3)
Differentiating again both sides w.r.t.𝑥
𝑒 𝑦 + 3𝑦 𝑦 𝑒 + 𝑒 𝑦 = 𝑒 ---- (4)
Differentiating again both sides w.r.t.𝑥

𝑒 𝑦 + 𝑦 𝑒 𝑦 + 3 𝑦 𝑦 𝑒 𝑦 + 𝑒 (𝑦 𝑦 + 𝑦 ) + 𝑒 3𝑦 𝑦 + 𝑦 𝑒 𝑦 = 𝑒 ---- (5)

Put 𝑥 = 0 in (1), (2), (3), (4) and (5)

(1) ⇒ (𝑦) = log 2
(2) ⇒ (𝑦 ) = =

(3) ⇒ 𝑒 𝑦 +𝑒 =1
1
(𝑦 ) =
4
(4) ⇒ 2𝑦 + 3 2+2 =1
3 1
2𝑦 + + = 1
4 4
(𝑦 ) = 0
(5) ⇒ (𝑦 ) = −

∴ By Maclaurin
Maclaurin's theorem
𝑥
𝑦 = (𝑦) + 𝑥 (𝑦 ) + (𝑦 ) + ⋯
2!
1 1 1
log(1 + 𝑒 ) = 𝑙𝑜𝑔2
𝑙𝑜𝑔 + 𝑥 + 𝑥 − 𝑥 +⋯
2 8 192

Can 𝑓(𝑥) = 𝑒 𝑥 ≠ 0 be expanded by Maclaurin’s theorem?

0 𝑥=0

Let 𝑓(𝑥) = 𝑒 ,𝑥 ≠ 0
Let us look into the differentiability of the function at 𝑥 = 0
𝑓(0 + ℎ) − 𝑓(0)
lim 𝑓 ((0)) = lim
→ → ℎ

𝑒
= lim
→ ℎ
Substituting = 𝜃, we can write

𝑒
lim 𝑓 (0) = lim
→ → 1/𝜃
𝜃
= lim
𝑒 →

1
= lim =0
→ 2𝜃𝑒

𝑓(0 − ℎ) − 𝑓(0)
lim 𝑓 ((0)) = lim
→ → −ℎ

𝑒
= lim =0
→ −ℎ
∴ 𝑓 (0) = 0
Also, 𝑓 (𝑥) = 𝑒 ,𝑥 ≠ 0
2
lim 𝑓 (𝑥) = lim 𝑒
→ → 𝑥
Substituting = 𝑡

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Calculus
2𝑡
lim 𝑓 (𝑥) = lim = 0 = 𝑓′(0)
→ → 𝑒
Therefore the function is continuous at 𝑥 = 0.
If we find the higher derivative of 𝑓(𝑥)for 𝑥 ≠ 0, we will get 𝑒 multiplied by a
polynomial in .
Therefore, higher derivatives of 𝑓(𝑥)will be zero at 𝑥 = 0.
So, the function possesses continuous derivatives for every value of 𝑥.
By Maclaurin’s theorem
𝑥 𝑥
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 (0) + 𝑓 (0) + ⋯ + 𝑓 (0) + 𝑅
2! (𝑛 − 1)!
𝑥 𝑥
and so𝑒 = 0 + 𝑥. 0 + . 0 + ⋯ .0 + 𝑅
2! (𝑛 − 1)!
i.e.𝑅 = 𝑒
𝑅 does not approach to zero as 𝑛 approaches to infinity.
Therefore, 𝑓(𝑥) can not be expanded by Maclaurin’s theorem

Summary
In this unit, we learnt about the finite form of Taylor’s and Maclaurin’s theorem.
 The Taylor’s theorem states that, if a function 𝑓 is such that
(i) the (𝑛 − 1) derivative 𝑓 is continuous in [𝑎, 𝑎 + ℎ],
(ii) the 𝑛 derivative 𝑓 exists in (𝑎, 𝑎 + ℎ) and
(iii) 𝑝 is a given positive integer
Then there exists at least one 𝜃 ∈ (0,1) such that
𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎)
𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) + (𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎)
1! 2! 𝑛!
(𝑥 − 𝑎) (1 − 𝜃)
+ 𝑓 (𝑎 + 𝜃(𝑥 − 𝑎))
(𝑛 − 1)! 𝑝
 The Taylor’s theorem with Lagrange’s form of remainder is given as,
𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎) ℎ
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ +⋯+ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2! (𝑛 − 1)! 𝑛!
 The Taylor’s theorem with Cauchy’s form of remainder is given as,
𝑓 (𝑎) 𝑓 (𝑎) 𝑓 (𝑎) ℎ (1 − 𝜃)
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ +ℎ +⋯+ℎ + 𝑓 (𝑎 + 𝜃ℎ)
1! 2! (𝑛 − 1)! (𝑛 − 1)!!
 For the Maclaurin’s theorem with Lagrange and Cauchy’s form of remainder,
substitute𝑎 = 0 and ℎ = 𝑥 in the above expressions.

Keywords
Taylor’s theorem, Maclaurin’s theorem, Lagrange’s form of remainder, Cauchy’s form of
remainder, Taylor’s series, Maclaurin’s series

Self Assessment
1. If a function 𝑓 is derivable then which of the following is true?
A. 𝑓 is defined
B. 𝑓 is defined and is continuous in a neighborhood of a point 𝑐
C. 𝑓 is defined and is uniformly continuous in a neighborhood of a point 𝑐
D. none of these

2. Expansion of function f(x) is?

A. 𝑓(0) + 𝑓 (0) + 𝑓 (0) + ⋯ + 𝑓 (0)
! ! !
B. 1+ 𝑓 (0) + 𝑓 (0) + ⋯ + 𝑓 (0)
! ! !
C. 𝑓(0) − 𝑓 (0) + 𝑓 (0) − ⋯ + 𝑓 (0)
! ! !
D. 𝑓(1) + !
𝑓 (1) + !
𝑓 (1) + ⋯ + !
𝑓 (1)

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Unit 10: Maclaurin’s and Taylor’s Theorems
3. The necessary condition for the Maclaurin expansion to be true for function f(x) is that
A. f(x) should be continuous
B. f(x) should be differentiable
C. f(x) should exist at every point
D. f(x) should be continuous and differentiable

4. The expansion of f(a+h) is

( )
A. 𝑓(𝑎) − !
𝑓 (a) + !
𝑓 (a) − ⋯ + !
𝑓 (a)
B. ℎ𝑓(𝑎) + ! 𝑓 (a) + !
𝑓 (a) + ⋯ + !
𝑓 (a)
C. 𝑓(ℎ) + !
𝑓 (h) + !
𝑓 (ℎ) + ⋯ + !
𝑓 (h)
D. 𝑓(𝑎) + 𝑓 (a) + 𝑓 (a) + ⋯ + 𝑓 (a)
! ! !

5. The expansion of 𝑒 is
A. 1 + + + +⋯
B. 1− + + +⋯
C. 1+ − + +⋯
D. 1 + + − +⋯

6. The (𝑛 + 1) term in the generalized mean value theorem or the Taylor theorem for the
function 𝑓(𝑎 + ℎ) is
( )
A. ( )!
𝑓 (𝑎 + 𝜃ℎ)
( )
B. ( )!
𝑓 (𝑎 + 𝜃ℎ)
( )
C. ( )!
𝑓 (𝑎 + 𝜃ℎ)
( )
D. ( )!
𝑓 (𝑎 − 𝜃ℎ)

( )
7. In the expression ( )!
𝑓 (𝑎 + 𝜃ℎ) which of the following is true for the 𝜃 value?
A. 𝜃 ∈ [0,1]
B. 𝜃 ∈ (0,1)
C. 𝜃 can take any value
D. 𝜃>0

( )
8. The expression ( )!
𝑓 (𝑎 + 𝜃ℎ) is known as
A. Remainder term
B. Remainder after n terms
C. Remainder after n+1 terms
D. Remainder after n-1 terms

( )
9. ( )!
𝑓 (𝑎 + 𝜃ℎ) is due to
A. Schlomilch
B. Lagrange
C. Schlomilch and Roche
D. Cauchy

10. 𝑓 (𝑎 + 𝜃ℎ) is the remainder after n terms due to

!
A. Schlomilch
B. Lagrange
C. Schlomilch and Roche
D. Cauchy

11. The Taylor’s theorem with Lagrange’s form of remainder for a function 𝑓(𝑥) will fail for
those values of 𝑥 for which
I. 𝑓(𝑥) or any of its differential coefficients becomes infinite
II. 𝑓(𝑥) or any of its differential coefficients becomes discontinuous
III. The remainder term is non-zero as 𝑛 → ∞

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Calculus
A. Only I is true
B. Only II is true
C. II and III are true
D. I, II and III are true

( )
12. The expression ( )!
𝑓 (𝑎 + 𝜃ℎ) is known as
A. Remainder term
B. Remainder after n terms
C. Remainder after n+1 terms
D. Remainder after n-1 terms

( )
13. ( )!
𝑓 (𝑎 + 𝜃ℎ) is due to
A. Schlomilch
B. Lagrange
C. Schlomilch and Roche
D. Cauchy

14. The function 𝑓(𝑥) = 𝑒 , 𝑥 ≠ 0 can be expanded by Maclaurin’s theorem.

0, 𝑥 = 0
A. True
B. False

15. The function 𝑓(𝑥) = √𝑥 can be expanded by Maclaurin’s theorem.

A. True
B. False

Answer for Self Assessment

1. B 2. A 3. D 4. D 5. D

6. C 7. B 8. B 9. C 10. B

11. D 12. B 13. D 14. B 15. B

Review Questions
1. Expand cos 𝑥 by Maclaurin’s series.
2. Expand log(1 + 𝑥) by Maclaurin’s theorem.
3. Expand log(𝑥 + 𝑎)in the powers of 𝑥 by Taylor’s theorem.
4. Expand log sin 𝑥 in powers of (𝑥 − 2).
5. Expand sin (𝑥 + ℎ) in powers of 𝑥 till the power of 𝑥 .
6. Expand tan 𝑥 in the powers of 𝑥 − .
7. Differentiate in the Taylor’s theorem with the Lagrange’s and Cauchy’s form of remainder.
8. Differentiate in the Maclaurin’s theorem with the Lagrange’s and Cauchy’s form of
remainder.
9. By Maclaurin’s theorem, find first three non vanishing terms in the expansion of .
10. Expand 𝑒 cos 𝑥 in the form of a power series.

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Unit 10: Maclaurin’s and Taylor’s Theorems

Further Reading
 W. Thomas Finny (1998). Calculus and Analytic Geometry, 66th Edition,
Publishers, Narsa, India.
 R. K. Jain, and Iyengar, SRK. (2010). Advanced Engineering Mathematics, 3 rd
Edition Publishers, Narsa, India.
 Widder, D.V. (2002). Advance Calculus 22nd Edition, Publishers, PHI, India.
 Piskunov, N. (1996). Differential and Integral Calculus Vol I, & II, Publishers,
CBS, India.

Online link
 https://openstax.org/books/calculus
https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-
maclaurin-series
 https://math.libretexts.org
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Ope
nStax)/10%3A_Power_Series/10.3%3A_Taylor_and_Maclaurin_Series

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 Notes
Dr. Pratiksha, Lovely Professional University Unit 11: Maxima and Minima of a Function

Unit 11: Maxima and Minima of a Function

CONTENTS
Objectives
Introduction
11.1 Absolute and Local Maximum / Minimum
11.2 A necessary Condition for Extreme Values
11.3 Sufficient Condition for Extreme Value
11.4 Second Order Derivative Test
Summary
Key Words
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will be able to

• derive necessary and sufficient condition for extreme values

• apply the first derivative test to find maxima and minima
• apply the second derivative test to calculate the maximum and minimum value of a
function

Introduction
In this unit, we will see one very interesting application of calculus, and it is called the maxima and
minima of a function. When one says, Mount Everest or Mariana Trench, what comes to your
mind? You think of a high point on the surface of Earth, and a low point on the surface of Earth. So
if you can draw the Earth's topography, the highest point will refer to a place which is a mountain,
and that gives you the idea of the maximum height for any object on the earth. Similarly, the
minimum height or you can say the maximum depth is at the Mariana Trench, so these ideas of
maximum and minimum are inherently there in our daily lives. We can see one more example to
understand the topic better.
The adjacent graph is about a cricket match between Australia and India. The blue one is
representing the run rate of India and the green one is representing the run rate of Australia, with
respect to the overs. By mathematical
modeling, we can write the run rate in
terms of overs, i.e. a function can be
framed or we can define some formula
in such a way that run rate, say ‘y’,
can be written in terms of ‘x’ where x
is the overs.
Here the graph is available, on the
basis of the data of the actual match.
In the first over the run rate for the
Australian team was two only. And in
the first over the run rate for Indian

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team was seven. In the very first over, India's run rate was maximum and then it fell down, and
then it rose up at a particular over (which one?), and then it rose then fell down then even more
down then it was constant for some time, and you can see the minimum run rate was at 11th over.
Just by looking at the graph it is quite clear, that for Australia, the minimum run rate was at the
second over, and the maximum was around at 10th over. So this is how, if we have a function we
can draw it and then from there, by just looking at the graph, we can tell about the maximum and
the minimum value of the function. But there should be a mathematical technique to deal with it,
without plotting the graphs! In this unit, we will look into how one can find the maximum and
minimum value of a function of one variable.
A high point is called a maximum (plural maxima).A low point is called a minimum (plural
minima).The general word for maximum or minimum is extremum (plural extreme).We say local
maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby.
Local maximum and minimum points are quite distinctive on the graph of a function, and are
therefore useful in understanding the shape of the graph. In many applied problems we want to
find the largest or smallest value that a function achieves (for example, we might want to find the
minimum cost at which some task can be performed) and so identifying maximum and minimum
points will be useful for applied problems as well.

11.1 Absolute and Local Maximum / Minimum

A function f has an absolute maximum (also called global maximum) at c if f (c) ≥ f (x) for all x in its
domain, D. The value f (c) is called the maximum value of f. A function f has an absolute minimum
(or global minimum) at c if f (c) ≤ f (x) for all x in its domain. Such a value f (c) is called the
minimum value of f. The maximum and minimum values of f are called the extreme values of f.
Whereas a function f has a local maximum (or relative maximum) at c if f (c) ≥ f (x) when x near c.
That is, f (c) ≥ f (x) for all x on some open interval containing c. Similarly, f has a local minimum (or
relative minimum) at c if f (c) ≤ f (x) when x near c.
Before proceeding, let’s note two important issues regarding this definition. First, the term absolute
here does not refer to absolute value. An absolute extremum may be positive, negative, or zero.
Second, if a function 𝑓has an absolute extremum over an interval 𝐼 at 𝑐the absolute extremum
is𝑓(𝑐) . The real number 𝑐is a point in the domain at which the absolute extremum occurs.
A function may have both an absolute maximum and an absolute minimum, just one extremum, or
neither. However, the following theorem, called the Extreme Value Theorem, guarantees that a
continuous function 𝑓(𝑥)over a closed, bounded interval [𝑎, 𝑏]has both an absolute maximum and
an absolute minimum.
The Extreme Value Theorem: If f is continuous on a closed interval [a, b] , then there exist (at least)
a point c where f attains its maximum value, f (c), on the interval, and (at least) a point d where f
attains its minimum value, f (d), on the interval.
This means that if both of the following conditions: (1) the interval is closed, and (2)f is continuous
on it, are met, then f is guaranteed to have (at least) one absolute maximum and one absolute
minimum points on the interval. If either condition fails, then the existence of max / min points is
not guaranteed.

11.2 A necessary Condition for Extreme Values

A necessary condition for 𝑓(𝑐) to be an extreme value of 𝑓 is that 𝑓 ′ (𝑐) = 0.
Let 𝑓(𝑐) be a maximum value of 𝑓. Then there exists an open interval (𝑐 − 𝛿, 𝑐 + 𝛿)around 𝑐, such
that if 𝑐 + ℎ is a number other than 𝑐 in (𝑐 − 𝛿, 𝑐 + 𝛿) , we have
𝑓(𝑐 + ℎ) < 𝑓(𝑐)
Here ℎ may be positive or negative. Thus
𝑓(𝑐 + ℎ) − 𝑓(𝑐)
ℎ>0 ⇒ <0
ℎ

𝑓(𝑐 + ℎ) − 𝑓(𝑐)
ℎ<0 ⇒ >0
ℎ

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Unit 11: Maxima and Minima of a Function
which implies that
𝑓(𝑐 + ℎ) − 𝑓(𝑐) 𝑓(𝑐 + ℎ) − 𝑓(𝑐)
lim+ ≤ 0 and lim− ≥0
ℎ→0 ℎ ℎ→0 ℎ
which will be true simultaneously if and only if 𝑓 ′ (𝑐) = 0.
Similarly the result holds if 𝑓(𝑐)a minimum value of is 𝑓.
Note that, 𝑓 ′ (𝑐) = 0 is not the sufficient condition for
𝑓(𝑐) to be an extreme value it can be explained with an
example.
Consider 𝑓(𝑥) = 𝑥 3 for 𝑥 = 0.
𝑓 ′ (𝑥) = 3𝑥 2
𝑓 ′ (0) = 0
Now 𝑥 > 0 ⇒ 𝑓(𝑥) > 0 = 𝑓(0)

𝑥 < 0 ⇒ 𝑓(𝑥) < 0 = 𝑓(0)

Therefore it can be seen that 𝑓(0) is not an extreme

value even though 𝑓 ′ (0) = 0.

Prove that the function 𝑓 defined by 𝑓(𝑥) = 3|𝑥| +

4|𝑥 − 1| ∀ 𝑥 ∈ 𝑅 has a minimum value 3 at 𝑥 = 1.
We can rewrite the function by using the definition of modulus function as follows:
4 − 7𝑥, 𝑥 < 0
4, 𝑥 = 0
𝑓(𝑥) = 4 − 𝑥, 0 < 𝑥 < 1
3, 𝑥 = 1
{ 7𝑥 − 4, 𝑥 > 1
Clearly it can be seen that at 𝑥 = 1, the function has a maximum value equal to 3.

A function is said to be stationary for 𝑐 and 𝑓(𝑐) a stationary value of 𝑓if 𝑓 ′ (𝑐) = 0. The rate of
change of a function is zero at the stationary point.

Find the greatest and least value of the function 𝑓(𝑥) = 3𝑥 4 − 2𝑥 3 − 6𝑥 2 + 6𝑥 + 1 in [0,2].
𝑓 ′ (𝑥) = 12𝑥 3 − 6𝑥 2 − 12𝑥 + 6
= 6(𝑥 − 1)(𝑥 + 1)(2𝑥 − 1)
Now by the necessary condition for an extreme value, 𝑓 ′ (𝑥) must be zero. And this gives 𝑥 =
1
1, −1, . Since -1 is not in the domain of the function, this value can be ignored. The other two
2
numbers are the candidates to be the point of maximum or minimum value of the function.
𝑓(1) = 2
1 39
𝑓( ) = = 2.43
2 16
Moreover we must check the value of the function at the end points of the domain also.
𝑓(0) = 1
𝑓(2) = 21
Thus the function has its maximum value at 𝑥 = 2 and minimum value at 𝑥 = 0.

11.3 Sufficient Condition for Extreme Value

𝑓(𝑐)is an extreme value of 𝑓 if and only if 𝑓′(𝑥) changes sign as 𝑥 passes through 𝑐.

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Fermat’s Theorem (test for local extreme values): If f has a local maximum or minimum at c, and if
f ′(c) exists, then f ′(c) = 0.
Note: Therefore, it follows that a local extreme point can only occur at places where either f ′(x) = 0
or f ′(x) is undefined (i.e., either at a point where the tangent line is horizontal, or at a non-
differentiable point). Examples: f (x) = (x − 2)2, at x = 2; g(x) =│x│, at x = 0.
Note: The converse is not always true: the fact that f ′(c) = 0, or that f ′(c) does not exist, does NOT
guarantee that c is a local extreme point of f. Example: f (x) = x3, at x = 0.
A critical point or critical number of a function f is a point x = c in the domain of f such that either
f ′(c) = 0 or f ′(c) does not exist.
Note: The critical points are all the candidate points for local maximum / minimum of f . That is,
every local extreme point is a critical point, but not every critical point is a local extreme point.
Naturally, the maximum/ minimum points of f have to be in the domain of f , i.e. they are points on
the graph of f. Therefore, for example, if f is undefined at an infinite discontinuity then the point of
discontinuity is not a critical point even though f ′ does not exist there.

Steps to find the absolute maximum and minimum values of a continuous function f
on a closed interval:
1. Find all critical points of f in the given interval.
2. Evaluate f at the critical point(s) found in step 1, as well as at the two endpoints of the interval.
3. The point(s) of the largest value of f is the absolute maximum(s), the point(s) of the smallest
value is the absolute minimum(s).
Let us understand this by an example.

Examine the polynomial 𝑓(𝑥) = 10𝑥 6 − 24𝑥 5 + 15𝑥 4 − 40𝑥 3 + 108 for maximum and
minimum value
Here 𝑓 ′ (𝑥) = 60𝑥 2 (𝑥 2 + 1)(𝑥 − 2)
For maximum and minimum value 𝑓 ′ (𝑥) = 0 implies that 𝑥 = 0, 2 are the only real values.
Now 𝑥 < 0 ⇒ 𝑓 ′ (𝑥) < 0

0 < 𝑥 < 2 ⇒ 𝑓 ′ (𝑥) < 0and

𝑥 > 2 ⇒ 𝑓 ′ (𝑥) > 0

Therefore 𝑓′(𝑥) does not change sign as 𝑥 passes through 0, so that 𝑓(0) is neither a maximum nor a
minimum value and 𝑓′(𝑥) changes the sign from negative to positive as 𝑥 passes through 2.
∴ 𝑓(2) = −100is the minimum value.
∴ 𝑓(𝑥)has only one extreme value i.e. at 2.

Find all local maximum and minimum points for the function
𝑓(𝑥) = 𝑥 3 − 𝑥
The derivative is𝑓 ′ (𝑥) = 3𝑥 2 − 1.
1
This is defined everywhere and is zero at 𝑥 = ± .
√3

1 1 2√3
Looking first at 𝑥 =
√3
. we see that 𝑓 ( ) = −
√3 9
.
1
Now we test two points on either side of 𝑥 = , making sure that neither is farther away than the
√3
1
nearest critical value; since √3 < 3, < 1 and we can use 𝑥 = 0 and 𝑥 = 1 .
√3

2√3 2√3 1
Since 𝑓(0) = 0 > − and𝑓(1) = 0 > − , there must be a local minimum at𝑥 = .
9 9 √3

1 1 2√3
For𝑥 = − ., we see that(− )= .
√3 √3 9

This time we can use 𝑥 = 0and 𝑥 = −1.

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Unit 11: Maxima and Minima of a Function
2√3 1
We find that(−1) = 𝑓(0) = 0 < , so there must be a local maximum at 𝑥 = − .
9 √3

This example is made very simple by our choice of points to test, for other choice of points the
calculations would have been comparatively lengthy.

11.4 Second Order Derivative Test

Theorem: 𝑓(𝑐) is a minimum value of the function 𝑓 if 𝑓 ′ (𝑐) = 0 and𝑓 ′′ (𝑐) > 0.
Proof:𝑓 ′′ (𝑐) > 0 ⇒ ∃ an open interval(𝑐 − 𝛿, 𝑐 + 𝛿) around 𝑐 for every point 𝑥 of which, the second
derivative is positive.
⇒ 𝑓 ′ (𝑥)is strictly increasing in (𝑐 − 𝛿, 𝑐 + 𝛿).

Also, 𝑓 ′ (𝑐) = 0
∴ 𝑓 ′ (𝑥) < 0 ∀ 𝑥 ∈ [𝑐 − 𝛿, 𝑐) (strictly decreasing function)
and 𝑓 ′ (𝑥) > 0 ∀ 𝑥 ∈ [𝑐, 𝑐 + 𝛿) (strictly increasing function)

⇒ 𝑓 (𝑐)is a minimum value of 𝑓(𝑥).

Theorem: 𝑓(𝑐) is a maximum value of the function 𝑓 if 𝑓 ′ (𝑐) = 0 and𝑓 ′′ (𝑐) < 0.
Proof: 𝑓 ′′ (𝑐) < 0 ⇒ ∃ an open interval(𝑐 − 𝛿, 𝑐 + 𝛿) around 𝑐 for every point 𝑥 of which, the second
derivative is negative.
⇒ 𝑓 ′ (𝑥)is strictly decreasing in (𝑐 − 𝛿, 𝑐 + 𝛿).

Also, 𝑓 ′ (𝑐) = 0
∴ 𝑓 ′ (𝑥) > 0 ∀ 𝑥 ∈ [𝑐 − 𝛿, 𝑐)(strictlyincreasing function)
and 𝑓 ′ (𝑥) < 0 ∀ 𝑥 ∈ [𝑐, 𝑐 + 𝛿) (strictly decreasing function)

⇒ 𝑓 (𝑐)is a maximum value of 𝑓(𝑥).

So, from now onwards we can follow the following steps to find the maximum / minimum of a
function:

Find the first derivative of the

function

Equate the first derivative to zero

and find all the critical points

Calculate the second order derivative

of the function

For each critical value, find the value of the

value of the second order derivative and
decide for maximum/minimum

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If the second derivative comes out to

be zero, then use the first derivative
test

1 𝑥 1
Show that the maximum value of ( ) is 𝑒 𝑒 .
𝑥

1 𝑥
Let 𝑦 = ( )
𝑥

log 𝑦 = −𝑥 𝑙𝑜𝑔𝑥
Differentiating both sides w.r.t.𝑥.
𝑑𝑦 1 𝑥
= −(1 + log 𝑥) ( )
𝑑𝑥 𝑥
Applying the necessary condition for extreme values, we get
log 𝑥 = −1
𝑥 = 𝑒 −1
Now by the sufficient condition, we can check if the point 𝑥 = 𝑒 −1 is the point of maximum or a
point of minimum or neither of them.
1
𝑑2𝑦
At = 𝑒 −1 , = −𝑒. 𝑒 𝑒 < 0
𝑑𝑥 2
1
Therefore 𝑦 has a maximum for 𝑥 = 𝑒 −1 and the maximum value is 𝑒 𝑒 .

Find the maximum and minimum value of the function𝑓(𝑥) = 8𝑥 5 − 15𝑥 4 + 10𝑥 2.
The given function is 𝑓(𝑥) = 8𝑥 5 − 15𝑥 4 + 10𝑥 2
𝑓 ′ (𝑥) = 40𝑥 4 − 60𝑥 3 + 20𝑥
= 20𝑥(2𝑥 3 − 3𝑥 2 + 1)
= 20𝑥(𝑥 − 1)2 (2𝑥 + 1)
Putting 𝑓 ′ (𝑥) = 0 for the critical points, we get
1
𝑥 = 0, 1, −
2
Now these three points are the candidates to be the point of maximum or minimum or neither of
them. Let’s check with the help of the second derivative test.
We have
𝑓 ′′ (𝑥) = 160𝑥 3 − 180𝑥 2 + 20
𝑓 ′′ (𝑥)(𝑎𝑡 𝑥 = 0) = 20 > 0 ⇒ 𝑥 = 0 is a point of minimum

𝑓 ′′ (𝑥)(𝑎𝑡 𝑥 = 1) = 0 ⇒ 𝑥 = 1 is neither a point of minimum nor of maximumbecause𝑓 ′ (𝑥) does not

change sign as 𝑥 passes through 1.
1 1
𝑓 ′′ (𝑥) (𝑎𝑡 𝑥 = − ) = −45 < 0 ⇒ 𝑥 = − is a point of maximum.
2 2

Find the absolute maximum and minimum points of f (x) = 4 − x2 on each of the intervals (i)
[−3, 1] and (ii) [2, 5].

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Unit 11: Maxima and Minima of a Function
f ′(x) = −2x
f ′ = 0 at x = 0 which is the only critical point because f is a polynomial, therefore, it has no non
differentiable points.
(i) Evaluate f at the critical point 0 and the endpoints −3 and 1:
f (−3) = −5, f (0) = 4, f (1) = 3
Therefore, the absolute maximum point is (0, 4), and the absolute minimum point is (−3, −5).
(ii) The critical point x = 0 is not in this interval, therefore, just evaluate f at the endpoints 2 and 5:
f (2) = 0, f (5) = −21
Therefore, the absolute maximum point is (2, 0), and the absolute minimum point is (5, −21).

Summary
In this unit we have seen how to calculate the maximum and minimum of a function if they exist.

• A function f has an absolute maximum (also called global maximum) at c if f (c) ≥ f (x) for
all x in its domain.
• A function f has an absolute minimum (or global minimum) at c if f (c) ≤ f (x) for all x in its
domain.
• The maximum and minimum values of f are called the extreme values of f.
• If f is continuous on a closed interval [a, b] , then there exist (at least) a point c where f
attains its maximum value, f (c), on the interval, and (at least) a point d where f attains its
minimum value, f (d), on the interval.
• 𝑓(𝑐) is an extreme value of 𝑓 if and only if 𝑓′(𝑥) changes sign as 𝑥 passes through 𝑐.
• 𝑓(𝑐)is a minimum value of the function 𝑓 if 𝑓 ′ (𝑐) = 0 and𝑓 ′′ (𝑐) > 0.
• 𝑓(𝑐)is a maximum value of the function 𝑓 if 𝑓 ′ (𝑐) = 0 and𝑓 ′′ (𝑐) < 0.

Key Words
Maxima, Minima, maximum of a function, minimum of a function, First derivative test, critical
points, stationary points, second derivative test, Extreme value theorem, Fermat’s theorem

Self Assessment
1. What is the saddle point?
A. Point where function has maximum value
B. Point where function has minimum value
C. Point where function has zero value
D. Point where function neither has maximum value nor minimum value

2. Which of the following is correct?

A. f(a) is an extreme value of f(x) if f '(a)=0
B. If f(a) is an extreme value of f(x), then f '(a) = 0
C. If f '(a) = 0, then f (a) is an extreme value of f(x)
D. All of these

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3. The maxima and minima of the function 𝑓(𝑥) = 2𝑥 3 − 15 𝑥 2 + 36𝑥 + 10 occur
respectively at
A. x= 3 and x = 2
B. x= 3 and x = 2
C. x = 2 and x = 3
D. x = 3 and x = 4

4. Find the maximum and minimum of 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 + 9𝑥 + 1 on the interval [0, 5].

A. maximum of f is 21 and the minimum is 1

B. maximum of f is 1 and the minimum is -21
C. maximum of f is 20 and the minimum is 1
D. maximum of f is 2 and the minimum is -11

5. A necessary condition for 𝑓(𝑐) to be an extreme value of 𝑓 is that

A. 𝑓 ′ (𝑐) ≠ 0
B. 𝑓(𝑐) = 0
C. 𝑓 ′ (𝑐) = 0
D. 𝑓 ′′ (𝑐) = 0

6. The maximum value of 𝑠𝑖𝑛𝑥 + cos 𝑥 is

A. 2
B. √2
C. 1
D. 1 + √2

1
7. The maximum value of 𝑓(𝑥) = 𝑥 3 − 2𝑥 2 + 3𝑥 + 1 is
3

A. 3/7
B. 7/3
C. 1
D. 7

8. 𝑓(𝑐)is a minimum value of the function if

A. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) > 0

B. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) < 0
C. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) = 0
D. 𝑓 ′′ (𝑐) > 0

9. 𝑓(𝑐)is a maximum value of the function if

A. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) > 0

B. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) < 0
C. 𝑓 ′ (𝑐) = 0 𝑎𝑛𝑑𝑓 ′′ (𝑐) = 0
D. 𝑓 ′ ′(𝑐) < 0
1 𝑥
10. The maximum value of ( ) is
𝑥

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Unit 11: Maxima and Minima of a Function
A. 𝑒
1
B. 𝑒𝑒
1 𝑒
C. ( )
𝑒

D. none of these

11. The saddle points for the function 10𝑥 6 − 24𝑥 5 + 15𝑥 4 − 40𝑥 3 + 108 are

A. 0, 1, 2
B. -1, 0, 2
C. 1, 2
D. 0, 2

12. The saddle points of the function 8𝑥 5 − 15 𝑥 4 + 10 𝑥 2 are given as

A. -1/2, 0
B. 0, 1, ½
C. -1/2, 0, 1
D. none of these

13. For the function 8𝑥 5 − 15 𝑥 4 + 10 𝑥 2 , 𝑥 = 0 is

A. a point of minimum
B. a point of maximum
C. point of inflexion
D. none of these

1
14. For the function8𝑥 5 − 15 𝑥 4 + 10 𝑥 2, 𝑥 = − is
2

A. a point of minimum
B. a point of maximum
C. point of inflexion
D. none of these

15. For the function8𝑥 5 − 15 𝑥 4 + 10 𝑥 2, 𝑥 = 1 is

A. a point of minimum
B. a point of maximum
C. point of inflexion
D. neither a point of maximum nor a point of minimum

Answer for Self Assessment

1 D 2. D 3. C 4. A 5. C

6. B 7. B 8. A 9. B 10. B

11. D 12. C 13. C 14. B 15. D

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Review Questions
1. Find the absolute maximum and minimum values of f (x) = x3 − 27x + 8 on the interval [0,
4].
2. Find the absolute maximum and minimum values of g(t) = t3/5 on the interval [−32, 1].
3. Find the maximum and minimum values of f (x) = x3 − 27x + 8.
4. Find the maximum and minimum values of g(t) = t3/5
5. Find the greatest and least value of the function 𝑥 4 − 4𝑥 3 − 2𝑥 2 + 12𝑥 + 1 in the
interval[−2, 5].
6. Find the greatest and least values of the function 2𝑥 3 − 15𝑥 2 + 36𝑥 + 1 in the interval
[2, 3] as well as in the interval[0, 4].
7. Show that the function 𝑓(𝑥) = (𝑥 + 2)(𝑥 − 1)2 (2𝑥 − 1)(𝑥 − 3) changes sign from positive
1
to negative as 𝑥 passes through and from negative to positive as 𝑥 passes through -2 or
2
3. Also show that it does not change sign as 𝑥 passes through 1.
8. Show that 𝑥 5 − 5𝑥 4 + 5𝑥 2 − 1 has a maximum value when 𝑥 = 1, a minimum value when
𝑥 = 3 and neither when 𝑥 = 0.
9. Show that the function𝑓 defined by
𝑓(𝑥) = 𝑥 𝑝 (1 − 𝑥)𝑞 ∀ 𝑥 ∈ 𝑹
where𝑝, 𝑞 are positive integers, has a maximum value for
𝑝
𝑥= ∀ 𝑝, 𝑞
𝑝+𝑞
10. Find the extreme value of the expression:
𝑥3
4
(𝑥 + 1)
11. Determine the value of 𝑥 for which
𝑥
1 + 𝑥𝑡𝑎𝑛𝑥
has a maximum value.
12. Find the maximum and minimum value of𝑠𝑖𝑛𝑥 𝑐𝑜𝑠2𝑥.
13. Show that a cubic polynomial can have at most two critical points. Give examples to show
that a cubic polynomial can have zero, one, or two critical points.
14. What can be said for a quadratic polynomial with respect to the critical points?

Further Reading
W. Thomas Finny (1998). Calculus and Analytic Geometry, 6th Edition, Publishers, Narsa,
India.
R. K. Jain, and Iyengar, SRK. (2010). Advanced Engineering Mathematics, 3 rd Edition
Publishers, Narsa, India.
Widder, D.V. (2002). Advance Calculus 2nd Edition, Publishers, PHI, India.
Piskunov, N. (1996). Differential and Integral Calculus Vol I, & II, Publishers, CBS, India.

https://opentextbc.ca/calculusv1openstax/chapter/maxima-and-minima/
https://www.mathsisfun.com/calculus/maxima-minima.html
https://www.whitman.edu/mathematics/calculus_online/section05.01.html

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 Notes
Dr. Pratiksha, Lovely Professional University Unit 12: Curvature and Asymptotes

Unit 12: Curvature and Asymptotes

CONTENTS
Objectives
Introduction
12.1 Curvature
12.2 Radius of Curvature
12.3 Length of arc as a Function
12.4 Radius of Curvature- Cartesian Equations
12.5 Radius of Curvature- Parametric Equations
12.6 Radius of Curvature- Polar Equations
12.7 Asymptotes of a General Algebraic Curve
12.8 Determination of Asymptotes
12.9 Asymptotes Parallel to the Coordinate Axis
12.10 Oblique Asymptotes
Summary
Key Words
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will be able to

• calculate the curvature for different types of curves

• distinguish between various kinds of asymptotes
• find the asymptotes of a general algebraic curve
• find the parallel and oblique asymptotes

Introduction
This unit is about two important
applications of derivatives namely
curvature and the asymptotes. Consider
that you are having a road trip in a hilly
region. Imagine the roads. You have the
technique to measure the distance
between any two points on a straight line.
But how to measure the bend happening
at a particular point needs some
elaboration on curvature! Similarly
another crucial feature of differential
calculus is the concept of asymptotes.
Basically it provides a frame for any curve

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Calculus
(if asymptotes exist). Both these concepts are very helpful to trace the curves.

12.1 Curvature
Curvature is the numerical measure of bending of a curve. At a particular point on the curve, a
tangent can be drawn. Let this line makes an angle 𝜓 with positive x- axis. Then curvature is
defined as the magnitude of rate of change of 𝜓with respect to the arc length 𝑠.

The total bending or total curvature is Arc𝑃𝑄 or Angle Δ𝜓

Δ𝜓
The average curvature is
Δ𝑠
Δ𝜓 d𝜓
The curvature of the curve at P is lim =
𝑄→𝑃 Δ𝑠 d𝑠

It is quite intuitive that the smaller circle bends more sharply than larger circle and thus smaller
circle has a larger curvature and larger the circle, smaller will be its curvature.
Let us consider a circle with center 𝑂 and radius 𝑟. Let the arc𝑃𝑄 = Δ𝑠

Arc 𝑃𝑄
Angle 𝑃𝑂𝑄 =
𝑂𝑃

Δ𝜓 1
=
Δ𝑠 𝑟
Δ𝜓 𝑑𝜓 1
lim = =
𝑄→𝑃 Δ𝑠 𝑑𝑠 𝑟
Therefore, curvature at any point of a circle is the reciprocalof the radius, and hence is a constant.

12.2 Radius of Curvature

The reciprocal of the curvature of a curve at any point in case it is non zero, is called its radius of
curvature at that point. It is denoted generally by Greek alphabet𝜌(rho).
𝑑𝑠
𝜌=
𝑑𝜓

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Unit 12: Curvature and Asymptotes
Note that the radius refers to the distance between the center of a circle and any other point on the
circumference of the circle. While the radius of curvature is the radius of the circle that touches the
curve at a given point. Also, it has the same tangent and curvature at that point.
The radius is of a real figure or shape whereas the radius of curvature is of an imaginary circle at a
point on a given curve.

Find the radius of curvature at any point for the curve 𝑠 = 𝑐 tan 𝜓
We have
𝑑𝑠
𝜌=
𝑑𝜓
∴ 𝜌 = 𝑐 sec 2 𝜓

Find the radius of curvature at any point for the curve 𝑠 = 𝑐 log sec 𝜓
We have
𝑑𝑠
𝜌=
𝑑𝜓
∴ 𝜌 = 𝑐 𝑡𝑎𝑛 𝜓

12.3 Length of arc as a Function

Let 𝑦 = 𝑓(𝑥) be the equation of the curve. 𝑃(𝑥, 𝑦)is any point on the curve such
that the arc length 𝐴𝑃 = 𝑠. The point 𝑄(𝑥 + Δ𝑥, 𝑦 + Δ𝑦) is a point near point P on
the curve.
Then arc 𝐴𝑄 = 𝑠 + Δ𝑠
So arc 𝑃𝑄 = Δ 𝑠
In triangle 𝑃𝑄𝑁, 𝑃𝑄 2 = (Δ𝑥)2 + (Δ𝑦)2
𝑃𝑄 2 Δ𝑦 2
( ) = 1+( )
Δ𝑥 Δ𝑥
On the left side introducing the arc 𝑃𝑄 in numerator and the denominator, we
get

chord𝑃𝑄 2 Δ𝑠 2 Δ𝑦 2
( ) ( ) = 1+( )
𝑎rc𝑃𝑄 Δ𝑥 Δ𝑥
As 𝑄 → 𝑃, the chord 𝑃𝑄 and arc 𝑃𝑄 become almost same making the above expression as

d𝑠 2 d𝑦 2
( ) =1+( )
d𝑥 d𝑥

d𝑠 d𝑦 2
= √1 + ( )
d𝑥 d𝑥

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Calculus
𝑑𝑥 𝑑𝑦
Corollary: 1. = cos 𝜓, = 𝑠𝑖𝑛𝜓
𝑑𝑠 𝑑𝑠

2. For parametric equations with parameter 𝑡, 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡)

d𝑠 2 d𝑥 2 d𝑦 2
( ) =( ) +( )
d𝑡 d𝑡 d𝑡
12.4 Radius of Curvature- Cartesian Equations
Consider a curve 𝑦 = 𝑓(𝑥)
𝑑𝑦
We have tan 𝜓 =
𝑑𝑥

Differentiating w.r.t. 𝑠 on both sides

𝑑𝜓 𝑑2 𝑦 𝑑𝑥
sec 2 𝜓 =
𝑑𝑠 𝑑𝑥 2 𝑑𝑠
3
𝑑𝑠 (1 + 𝑦12 )2
=𝜌=
𝑑𝜓 𝑦2
If 𝑦2 > 0, 𝜌 > 0 at a point, then the curve will be concave upward at that point.
If 𝑦2 < 0, 𝜌 < 0 at a point, then the curve will be concave downward at that point.
∵ 𝜌is independent of the choice of x-axis and y-axis ∴ 𝜌 can also be given as
3
𝑑𝑥 2 2
[1 + ( ) ]
𝑑𝑦
𝜌= 𝑑2𝑥
𝑑𝑦 2

3𝑎 3𝑎 8√2
Show that the curvature of the point ( , ) on the Folium 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦is − .
2 2 3𝑎

Differentiation the equation of folium w.r.t. 𝑥, we get

𝑑𝑦 𝑎𝑦 − 𝑥 2
=
𝑑𝑥 𝑦 2 − 𝑎𝑥
𝑑𝑦
( )3𝑎 3𝑎 = −1
𝑑𝑥 ,
2 2

And
𝑑2 𝑦 32
( 2) =−
𝑑𝑥 3𝑎,3𝑎 3𝑎
2 2

3𝑎 3𝑎
Therefore the curvature at the point ( , ) is
2 2
3
𝑑𝑥 2 2

1 [1 + ( ) ]
𝑑𝑦
= 𝑑2𝑥
𝜌
𝑑𝑦 2

8√2
= −
3𝑎

12.5 Radius of Curvature- Parametric Equations

For a curve given by 𝑥 = 𝑓(𝑡), 𝑦 = 𝑔(𝑡), For 𝑓 ′ (𝑡) ≠ 0
𝑑𝑦
𝑑𝑦 𝑑𝑡 𝑔′ (𝑡)
= 𝑑𝑥
=
𝑑𝑥 𝑓 ′ (𝑡)
𝑑𝑡
2 ′ (𝑡)𝑔′′ (𝑡)
𝑑 𝑦 1 𝑓 − 𝑔′ (𝑡)𝑓 ′′ (𝑡)
= ( )
𝑑𝑥 2 𝑓 ′ (𝑡) 𝑓 ′ (𝑡)2

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Unit 12: Curvature and Asymptotes
3
(𝑓 ′2 (𝑡) + 𝑔′2 (𝑡)) 2
∴𝜌=
𝑓 ′ (𝑡)𝑔′′ (𝑡) − 𝑔′ (𝑡)𝑓 ′′ (𝑡)

𝑡
For the cycloid 𝑥 = 𝑎(𝑡 + 𝑠𝑖𝑛𝑡), 𝑦 = 𝑎(1 − cos 𝑡), prove that 𝜌 = 4 a cos .
2

𝑑𝑥
= 𝑎(1 + cos 𝑡)
𝑑𝑡
𝑑𝑦
= 𝑎 sin 𝑡
𝑑𝑡
𝑑𝑦 𝑡
= tan
𝑑𝑥 2
𝑑2 𝑦 1 𝑡 𝑑𝑡
= sec 2
𝑑𝑥 2 2 2 𝑑𝑥
1 1
=
4𝑎 cos4 𝑡
2
3
𝑑𝑥 2 2
[1+( ) ] 𝑡
𝑑𝑦
The radius of curvature 𝜌 = 𝑑2 𝑥
= 4𝑎 cos .
2
𝑑𝑦2

12.6 Radius of Curvature- Polar Equations

Let 𝑟 = 𝑓(𝜃) be the given curve.
Let 𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃 be the transformations.
𝑑𝑥 𝑑𝑦
= 𝑟1 𝑐𝑜𝑠𝜃 − 𝑟 sin 𝜃, = 𝑟1 sin 𝜃 + 𝑟 cos 𝜃
𝑑𝜃 𝑑𝜃

𝑑𝑦 𝑟1 sin 𝜃 + 𝑟 cos 𝜃
∴ 𝑦1 = =
𝑑𝑥 𝑟1 𝑐𝑜𝑠𝜃 − 𝑟 sin 𝜃
3
𝑑2 𝑦 (𝑟 2 + 𝑟12 )2
𝑦2 = 2 =
𝑑𝑥 (𝑟1 𝑐𝑜𝑠𝜃 − 𝑟𝑠𝑖𝑛𝜃)3
where𝑟1 = 𝑓 ′ (𝜃), 𝑟2 = 𝑓′′(𝜃)
Thus the radius of curvature can be written as
3
(𝑟 2 + 𝑟12 )2
𝜌=
𝑟 2 + 2𝑟12 − 𝑟 𝑟2

𝑎𝑚
For the curve 𝑟 𝑚 = 𝑎𝑚 cos 𝑚𝜃, prove that 𝜌 = (𝑚+1)𝑟 𝑚−1.

Taking logarithm on both sides of the given equation and then differentiating w.r.t. 𝜃, we get
𝑚 𝑑𝑟 𝑠𝑖𝑛 𝑚𝜃
= −𝑚
𝑟 𝑑𝜃 𝑐𝑜𝑠𝑚𝜃
𝑟1 = −𝑟 tan 𝑚𝜃
𝑑2 𝑟
𝑟2 = = −𝑟𝑚 𝑠𝑒𝑐 2 𝑚𝜃 + 𝑟 tan2 𝑚𝜃
𝑑𝜃 2

Using
3
(𝑟 2 + 𝑟12 )2
𝜌=
𝑟 2 + 2𝑟12 − 𝑟 𝑟2
We get
𝜌 = (𝑟 3 sec 3 𝑚𝜃) /(𝑟 2 sec 2 𝑚𝜃 + 𝑚𝑟 2 sec 2 𝑚𝜃)
1 𝑎𝑚
𝜌=
𝑚 + 1 𝑟 𝑚−1

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Hence the result!

In 1643, French mathematician Rene Descartes developed a

formula relating the curvatures of four circles that all touch, or are
tangent, to each other.
c
Descartes’ Circle Equation Theorem:
Given four mutually tangent circles with curvatures a, b, c, and d
d, then b

(a2 + b2 + c2 + d2) = (1/2) (a + b + c + d)2 a

12.7 Asymptotes of a General Algebraic Curve

The name ‘asymptote’ originated from Greek word asymptotes which means ‘not meeting’.An
asymptote of a curve is a straight line such that the distance between the curve and the line
approaches to zero as one or both of the x or y coordinates tend to infinity. Simply put,an
asymptote is a line that a graph approaches without touching.
In some case a curve may have a branch or branches extending beyond the finite region. In this case
let P be a point on such a branch of the curve, having its coordinates (x,y) and if P moves along the
curve, so that at least one of x and y tend to + ∞ or to -∞, then P is said to tend to infinity.
A straight line is said to be an asymptote of a curve y = f (x), if the perpendicular distance of the
point P (x,y) on the curve from the line tends to 0 when x or y or both tend to infinity.
An asymptote parallel to y-axis may be referred as a vertical asymptote and parallel to x-axis as a
horizontal asymptote. An asymptote which is not parallel to either axis may be described as an
oblique asymptote and isgiven by y = mx + c. Only open curves which have some infinite branch
can have an asymptote. No closed curve can have an asymptote.The curve and its asymptote get
infinitely close, but they never meet.
Their major applications involve their usage in big O notation, they are simple approximations to
complex equations, and they are useful for graphing rational equations. In most cases, the
asymptote(s) of a curve can be found by taking the limit of a value where the function is not
defined.
For example, a cissoids given by the equation 𝑦 2 (2 − 𝑥) = 𝑥 3 , can be drawn as given below. It is
clear that the infinite branches of the curve seem to meet the straight line 𝑥 = 2.

12.8 Determination of Asymptotes

The general equation of a straight line is

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Unit 12: Curvature and Asymptotes
𝑦 = 𝑚𝑥 + 𝑐
Consider a point 𝑃 on the curve and let 𝑀 be the foot of the
perpendicular from point 𝑃 on the straight line that can be an
asymptote. As P(x, y) → ∞, x → ∞
The equation 𝑦 = 𝑚𝑥 + 𝑐 can be an asymptote of given curve if
𝑝 = 𝑃𝑀 and 𝑃𝑀 ⊥ 𝑁𝑀. Then
|𝑦 − 𝑚𝑥 − 𝑐|
𝑝=
√1 + 𝑚2
Now 𝑝 → 0 as 𝑥 → ∞
∴ lim 𝑦 − 𝑚𝑥 − 𝑐 = 0
𝑥→∞

𝐥𝐢𝐦 𝒚 − 𝒎𝒙 = 𝒄
𝒙→∞
𝑦 1
Also − 𝑚 = (𝑦 − 𝑚𝑥)
𝑥 𝑥

𝑦 1
lim ( − 𝑚) = lim (𝑦 − 𝑚𝑥). lim = 0
𝑥→∞ 𝑥 𝑥→∞ 𝑥→∞ 𝑥

This implies
𝒚
𝐥𝐢𝐦 =𝒎
𝒙→∞ 𝒙

Thus knowing 𝑚 and 𝑐, we can write 𝑦 = 𝑚𝑥 + 𝑐 as the equation of asymptote .

Examine the folium for asymptotes.

The folium is given by the equation 𝑥 3 + 𝑦 3 − 3𝑎𝑥𝑦 = 0 ---- (1)
𝑦 3 𝑦1
or1 + ( ) − 3𝑎 =0
𝑥 𝑥𝑥
---- (2)
We will calculate the slope and the intercept in the general
equation of a straight line under the definition of asymptote as
derived in the section above.
𝒚
Let 𝑥 → ∞ then 𝐥𝐢𝐦 = 𝒎
𝒙→∞ 𝒙

From equation (2) we can write 1 + 𝑚3 = 0

𝑚 = −1 is the only real root. For this value of 𝑚, we can find the
associated 𝑐.
𝑐 = lim (𝑦 + 𝑥)
𝑥→∞

Put 𝑦 + 𝑥 = 𝑝
As 𝑥 → ∞, 𝑝 → 𝑐
Put 𝑦 = 𝑝 − 𝑥 in (1)
𝑥 3 + (𝑝 − 𝑥)3 − 3𝑎𝑥(𝑝 − 𝑥) = 0
1 𝑝3
3(𝑝 + 𝑎) − 3𝑝(𝑝 + 𝑎) + 2 = 0
𝑥 𝑥
As 𝑥 → ∞, 𝑝 → 𝑐
3(𝑐 + 𝑎) = 0
𝑐 = −𝑎
∴ The equation of the asymptote is 𝑦 = −𝑥 − 𝑎.

12.9 Asymptotes Parallel to the Coordinate Axis

For asymptote parallel to the y axis:

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Calculus
Let 𝑥 = 𝑘 ----- (1)
be the asymptote of the curve. As 𝑃(𝑥, 𝑦) → ∞ along the curve 𝑦 → ∞ and 𝑃𝑀 = 𝑥 − 𝑘
lim 𝑥 − 𝑘 = 0
𝑦→∞

lim 𝑥 = 𝑘which gives 𝑘.

𝑦→∞

For a general formula we arrange the given curve in descending powers of 𝑦 i.e.
𝑦 𝑚 𝜙(𝑥) + 𝑦 𝑚−1 𝜙1 (𝑥) + 𝑦 𝑚−2 𝜙2 (𝑥) + ⋯ = 0 ---- (2)
where𝜙(𝑥), 𝜙1 (𝑥), 𝜙2 (𝑥) … are the polynomials in 𝑥.
Dividing (2) by 𝑦 𝑚 , we get
1 1
𝜙(𝑥) + 𝜙1 (𝑥) + 𝜙2 (𝑥) + ⋯ = 0 ---- (3)
𝑦 𝑦2

Let 𝑦 → ∞
then lim 𝑥 = 𝑘
𝑦→∞

𝜙(𝑘) = 0
Therefore 𝑘 is the root of equation 𝜙(𝑥) = 0. Let 𝑘1 , 𝑘2 etc. be the roots of 𝜙(𝑥) = 0, then the
asymptote parallel to y-axis are 𝑥 = 𝑘1 , 𝑥 = 𝑘2 etc.
⇒ (𝑥 − 𝑘1 )(𝑥 − 𝑘2 )etc. are the factors of 𝜙(𝑥) which is the coefficient of highest power 𝑦 𝑚 of y in the
given equation.
Similarly the derivation can be done for the asymptotes parallel to the x axis and can be
summarized as the following rules:
Rule 1: The asymptotes parallel to Y axis are obtained by equating to zero, the real linear factors in
the coefficient of highest power of y, in the equation of the curve.
Rule 2: The asymptotes parallel to X axis are obtained by equating to zero, the real linear factors in
the coefficient of highest power of x, in the equation of the curve.
Find the asymptote parallel to the coordinate axes of the curve
(𝑥 2 + 𝑦 2 )𝑥 − 𝑎𝑦 2 = 0
The equation can be re written as
𝑥 3 + 𝑦 2 (𝑥 − 𝑎) = 0
Coefficient of highest power of 𝑥 is 1 and that cannot be equated to zero. Therefore the asymptote
parallel to the x-axis does not exist.
Coefficient of highest power of 𝑦 is 𝑥 − 𝑎 and equating it to zero gives 𝑥 − 𝑎 = 0.Therefore the
asymptote parallel to the y-axis is 𝑥 = 𝑎.

12.10 Oblique Asymptotes

Asymptotes of the general rational algebraic equation
Consider the equation
𝑈𝑛 + 𝑈𝑛−1 + 𝑈𝑛−2 + ⋯ + 𝑈2 + 𝑈1 + 𝑈0 = 0 ---- (1)
𝑦 𝑦
where𝑈𝑟 is a homogeneous expression of degree 𝑟 in 𝑥, 𝑦 and 𝑈𝑟 = 𝑥 𝑟 𝜙𝑟 ( ) where 𝜙𝑟 ( ) is a
𝑥 𝑥
𝑦
polynomial in of degree 𝑟, at the most.
𝑥

Therefore (1) can be written as

𝑦 𝑦 𝑦 𝑦
𝑥 𝑛 𝜙𝑛 ( ) + 𝑥 𝑛−1 𝜙𝑛−1 ( ) + ⋯ + 𝑥𝜙1 ( ) + 𝜙0 ( ) = 0 ---- (2)
𝑥 𝑥 𝑥 𝑥

Dividing by 𝑥 𝑛
𝑦 1 𝑦 1 𝑦 1 𝑦
𝜙𝑛 ( ) + 𝜙𝑛−1 ( ) + ⋯ + 𝑛−1 𝜙1 ( ) + 𝑛 𝜙0 ( ) = 0
𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥
and taking the limit 𝑥 → ∞, we get

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Unit 12: Curvature and Asymptotes
𝜙𝑛 (𝑚) = 0 ---- (3)
which determines the slope of the asymptote.
Let 𝑚1 be one of the roots of this equation so that 𝜙𝑛 (𝑚1 ) = 0
We can write 𝑦 − 𝑚1 𝑥 = 𝑝1
𝑦 𝑝1
i.e. = 𝑚1 +
𝑥 𝑥

Substituting this in (2), we get

𝑝1 𝑝1 𝑝1 𝑝1
𝑥 𝑛 𝜙𝑛 (𝑚1 + ) + 𝑥 𝑛−1 𝜙𝑛−1 (𝑚1 + ) + ⋯ + 𝑥𝜙1 (𝑚1 + ) + 𝜙0 (𝑚1 + )=0
𝑥 𝑥 𝑥 𝑥

Expanding each term by Taylor’s theorem and re arranging the terms, we get

𝑝12 ′′ ′
𝑥 𝑛 𝜙𝑛 (𝑚1 ) + 𝑥 𝑛−1 (𝑝1 𝜙𝑛′ (𝑚1 ) + 𝜙𝑛−1 (𝑚1 )) + 𝑥 𝑛−2 ( 𝜙 (𝑚 ) + 𝑝1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 )) + ⋯ = 0
2 𝑛 1

Putting 𝜙𝑛 (𝑚1 ) = 0 and dividing by 𝑥 𝑛−1 , we get

1 𝑝12 ′
(𝑝1 𝜙𝑛′ (𝑚1 ) + 𝜙𝑛−1 (𝑚1 )) + ( 𝜙𝑛′′ (𝑚1 ) + 𝑝1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 )) + ⋯ = 0
𝑥 2

Let 𝑥 → ∞, we write lim 𝑝1 = 𝑐1

∴ 𝑐1 𝜙𝑛′ (𝑚1 ) + 𝜙𝑛−1 (𝑚1 ) = 0
𝜙𝑛−1 (𝑚1 )
or𝑐1 = − ′ (𝑚 ) , provided 𝜙𝑛′ (𝑚1 )≠0
𝜙𝑛 1

𝜙𝑛−1 (𝑚1 )
Therefore 𝑦 = 𝑚1 𝑥 − ′ (𝑚 ) is the asymptote corresponding to slope𝑚1 .
𝜙𝑛 1

𝜙𝑛−1 (𝑚2 ) 𝜙𝑛−1 (𝑚3 )

Similarly, 𝑦 = 𝑚2 𝑥 − ′ (𝑚 ) , 𝑦 = 𝑚3 𝑥 − ′ (𝑚 ) etc. are the asymptotes corresponding to 𝑚2 , 𝑚3
𝜙𝑛 2 𝜙𝑛 3
etc. which are the roots of 𝜙𝑛 (𝑚) = 0, such that the denominator of the fractions is non zero.
When 𝜙𝑛′ (𝑚1 ) = 0 and 𝜙𝑛−1 (𝑚1 ) ≠ 0
There does not exist any value of 𝑐1 . So there is no asymptote corresponding to the slope 𝑚1 .
Now suppose that 𝜙𝑛′ (𝑚1 ) = 0 = 𝜙𝑛−1 (𝑚1 ), then we can write
𝑝12 ′′ ′
1
𝜙 (𝑚 ) + 𝑝1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 ) + (… ) + ⋯ = 0
2 𝑛 1 𝑥
Taking the limit as 𝑥 → ∞, we get 𝑐1 is a root of the equation,
𝑐12 ′′ ′
𝜙 (𝑚 ) + 𝑐1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 ) = 0
2 𝑛 1
which determines two values of 𝑐1 say 𝑐1′ and 𝑐1′′ , provided that 𝜙𝑛′′ (𝑚1 ) ≠ 0.
Therefore 𝑦 = 𝑚1 𝑥 + 𝑐1′
and𝑦 = 𝑚1 𝑥 + 𝑐1′′ are the two asymptotes corresponding to the slope 𝑚1 and this is also known as
the case of parallel asymptotes.

Steps to find oblique asymptotes:

1. Put 𝑥 = 1, 𝑦 = 𝑚 in the highest degree term to get 𝜙𝑛 (𝑚) .

2. Similarly find 𝜙𝑛−1 (𝑚), 𝜙𝑛−2 (𝑚) etc.
3. Equate 𝜙𝑛 (𝑚) = 0 , solve for the real values of 𝑚.
𝜙𝑛−1 (𝑚1 )
4. Find the corresponding value of intercept say, 𝑐1 for the slope 𝑚1 using 𝑐1 = − ′ (𝑚 )
𝜙𝑛 1

5. Then 𝑦 = 𝑚𝑥 + 𝑐 is the required asymptote.

6. For the case of 𝜙𝑛′ (𝑚1 ) = 0 = 𝜙𝑛−1 (𝑚1 ), find 𝑐1 by the relation
𝑐12 ′′ ′
𝜙 (𝑚 ) + 𝑐1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 ) = 0
2 𝑛 1

Find the oblique asymptotes of the curve

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Notes
Calculus
2𝑥 3 − 𝑥 2 𝑦 − 2𝑥𝑦 2 + 𝑦 3 − 4𝑥 2 + 8𝑥𝑦 − 4𝑥 + 1 = 0
Attempting the first and second step; put 𝑥 = 1, 𝑦 = 𝑚 in the highest degree term to get 𝜙𝑛 (𝑚) , we
get,
𝜙3 (𝑚) = 2 − 𝑚 − 2𝑚2 + 𝑚3
𝜙2 (𝑚) = −4 + 8𝑚
𝜙1 (𝑚) = −4

Equate 𝜙3 (𝑚) = 0 , solve for the real values of 𝑚.

𝑚 = −1, 1, 2
𝜙𝑛−1 (𝑚 ) 𝜙2 (𝑚)
When 𝑚 = 2, 𝑐 = − ′ (𝑚 ) = − = −4
𝜙𝑛 𝜙3′ (𝑚)
The asymptote is 𝑦 = 2𝑥 − 4
𝜙𝑛−1 (𝑚 ) 𝜙2 (𝑚)
When 𝑚 = 1, 𝑐 = − ′ (𝑚 ) = − = 2
𝜙𝑛 𝜙3′ (𝑚)
The asymptote is 𝑦 = 𝑥 + 2

Find the asymptotes of 𝑥 3 − 𝑥 2 𝑦 − 𝑥𝑦 2 + 𝑦 3 + 2𝑥 2 − 4𝑦 2 + 2𝑥𝑦 + 𝑥 + 𝑦 + 1 = 0

𝜙3 (𝑚) = 1 − 𝑚 − 𝑚2 + 𝑚3
𝜙3′ (𝑚) = −1 − 2𝑚 + 3𝑚2
𝜙3′′ (𝑚) = −2 + 6𝑚
𝜙2 (𝑚) = 2 − 4𝑚2 + 2𝑚
𝜙2′ (𝑚) = −8𝑚 + 2
𝜙3 (𝑚) = 1 + 𝑚
⇒ 𝑚3 − 𝑚2 − 𝑚 + 1 = 0

(𝑚2 − 1)(𝑚 − 1) = 0
𝑚 = 1, 1, −1
𝜙 (𝑚)
When 𝑚 = 1, 𝑐 = − 2′ (𝑚)
𝜙3

𝜙2 (1) = 0, 𝜙3′ (1) = 0

𝑐 2 ′′
𝜙 (𝑚) + 𝑐𝜙2′ (𝑚) + 𝜙1 (𝑚) = 0
2 3
3 ± √5
𝑐=
2
3+√5 3−√5
The asymptotes corresponding to 𝑚 = 1 are 𝑦 = 𝑥 + and 𝑦 = 𝑥 +
2 2

Summary
• The total bending or total curvature is Arc𝑃𝑄 or Angle Δ𝜓
Δ𝜓
• The average curvature is
Δ𝑠
Δ𝜓 d𝜓
• The curvature of the curve at P is lim =
𝑄→𝑃 Δ𝑠 d𝑠
3
𝑑𝑥 2 2
[1+( ) ]
•
𝑑𝑦
Radius of curvature for a Cartesian curve 𝜌 = 𝑑2 𝑥
𝑑𝑦2
3
(𝑓′2 (𝑡)+𝑔′2 (𝑡))2
• Radius of curvature for a parametric curve 𝜌 =
𝑓′ (𝑡)𝑔′′ (𝑡)−𝑔′ (𝑡)𝑓′′ (𝑡)
3
(𝑟 2 +𝑟12)2
• Radius of curvature for a polar curve 𝜌 = 2 2
𝑟 +2𝑟1 −𝑟 𝑟2
• The asymptotes parallel to Y axis are obtained by equating to zero, the real linear factors in
the coefficient of highest power of y, in the equation of the curve.

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Unit 12: Curvature and Asymptotes
• The asymptotes parallel to X axis are obtained by equating to zero, the real linear factors in
the coefficient of highest power of x, in the equation of the curve.

• Steps to find oblique asymptotes:

1. Put 𝑥 = 1, 𝑦 = 𝑚 in the highest degree term to get 𝜙𝑛 (𝑚) .
2. Similarly find 𝜙𝑛−1 (𝑚), 𝜙𝑛−2 (𝑚) etc.
3. Equate 𝜙𝑛 (𝑚) = 0 , solve for the real values of 𝑚.
𝜙𝑛−1 (𝑚1 )
4. Find the corresponding value of intercept say, 𝑐1 for the slope 𝑚1 using 𝑐1 = − ′ (𝑚 )
𝜙𝑛 1

5. Then 𝑦 = 𝑚𝑥 + 𝑐 is the required asymptote.

6. For the case of 𝜙𝑛′ (𝑚1 ) = 0 = 𝜙𝑛−1 (𝑚1 ), find 𝑐1 by the relation
𝑐12 ′′ ′
𝜙 (𝑚 ) + 𝑐1 𝜙𝑛−1 (𝑚1 ) + 𝜙𝑛−2 (𝑚1 ) = 0
2 𝑛 1

Key Words
Curvature, radius of curvature, vertical asymptote, horizontal asymptote, oblique asymptotes

Self Assessment
1. The angle through which the tangent turns as a point moves along the curve from a point
P to Q, will be large or small as compared to arc length, depends upon
A. slope of tangent
B. sharpness of bend
C. velocity
D. acceleration

2. Which of the following is true?

I. The curvature of a circle is the same at every point.
II. Larger the circle, smaller will be its curvature.
A. Only I
B. Only II
C. Both I and II
D. None is true

3. The reciprocal of the curvature of a curve at any point in case it is non- zero, is called
A. curvature
B. radius of curvature
C. bend
D. total bending

4. The radius of curvature at any point for the curve 𝑠 = 𝑐 𝑡𝑎𝑛𝜓 is

A. 𝑐 sec 2 𝜓
B. 𝑐 sec 𝜓
C. 𝑐 sec 3 𝜓
D. 𝑐 cot 2 𝜓

5. For a curve if the radius of curvature is negative, it means that

A. the curve is concave upwards
B. the curve is concave downwards
C. the curve has no bend
D. none of these

6. For the cycloid 𝑥 = 𝑎(𝑡 + 𝑠𝑖𝑛 𝑡), 𝑦 = 𝑎(1 − 𝑐𝑜𝑠 𝑡), the radius of curvature is given as
A. 4𝑎 cos 𝑡
𝑡
B. 4 cos
2

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Notes
Calculus
C. 4 cos 𝑡
𝑡
D. 4𝑎 cos
2

7. Which of the following is true about asymptotes?

I. An asymptote of a curve is a line to which the curve converges.

II. The curve and its asymptote get infinitely close, but they never meet.
A. Only I is true
B. Only II is true
C. Both I and II are true
D. None is true

8. For the curve 𝑦 2 (2 − 𝑥) = 𝑥 3

A. there is no asymptote
B. there exists one asymptote only
C. there are two asymptotes
D. there are three asymptotes

9. The curve 𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0 has

A. one oblique asymptote
B. two oblique asymptotes
C. no asymptote
D. no oblique asymptotes

10. For the curve 𝑦 2 (2 − 𝑥) = 𝑥 3

A. there is one asymptote parallel to x-axis
B. there are two asymptotes parallel to y-axis
C. there are no asymptotes
D. there exists one asymptote parallel to y-axis

11. The curve 𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0 has

A. one parallel asymptote
B. two parallel asymptotes
C. no asymptote
D. no parallel asymptotes

12. For the curve 𝑥 2 𝑦 − 3𝑥 2 − 5𝑥𝑦 + 6𝑦 + 2 = 0, there are

A. one horizontal and two vertical asymptotes
B. two horizontal and two vertical asymptotes
C. two horizontal and one vertical asymptotes
D. one horizontal and one vertical asymptotes

13. A closed curve has

A. no asymptote
B. one asymptote
C. infinitely many asymptotes
D. n asymptotes where n is the degree of the curve

14. The asymptotes of the curve 𝑥𝑦 − 2𝑦 − 3𝑥 = 0 are given by

A. 𝑥 − 2 = 0, 𝑦 + 3 = 0
B. 𝑥 + 2 = 0, 𝑦 + 3 = 0
C. 𝑥 − 2 = 0, 𝑦 − 3 = 0
D. 𝑥 − 2 = 0, 𝑦 = 0

15. The asymptotes of the curve 𝑥 2 𝑦 2 − 𝑎2 (𝑥 2 + 𝑦 2 ) = 0 form a

A. circle
B. square
C. pentagon

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 Notes
Unit 12: Curvature and Asymptotes
D. triangle

16. Which of the following is true about asymptotes?

I. An asymptote of a curve is a line to which the curve converges.

II. The curve and its asymptote get infinitely close, but they never meet.
A. Only I is true
B. Only II is true
C. Both I and II are true
D. None is true

17. For the curve 𝑦 2 (2 − 𝑥) = 𝑥 3

A. there is no asymptote
B. there exists one asymptote only
C. there are two asymptotes
D. there are three asymptotes

18. The curve 𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0 has

A. one oblique asymptote
B. two oblique asymptotes
C. no asymptote
D. no oblique asymptotes

19. For the curve 𝑦 2 (2 − 𝑥) = 𝑥 3

A. there is one asymptote parallel to x-axis
B. there are two asymptotes parallel to y-axis
C. there are no asymptotes
D. there exists one asymptote parallel to y-axis

20. The curve 𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0 has

A. one parallel asymptote
B. two parallel asymptotes
C. no asymptote
D. no parallel asymptotes

21. For the curve 𝑥 2 𝑦 − 3𝑥 2 − 5𝑥𝑦 + 6𝑦 + 2 = 0, there are

A. one horizontal and two vertical asymptotes
B. two horizontal and two vertical asymptotes
C. two horizontal and one vertical asymptotes
D. one horizontal and one vertical asymptotes

22. A closed curve has

A. no asymptote
B. one asymptote
C. infinitely many asymptotes
D. n asymptotes where n is the degree of the curve

23. The asymptotes of the curve 𝑥𝑦 − 2𝑦 − 3𝑥 = 0 are given by

A. 𝑥 − 2 = 0, 𝑦 + 3 = 0
B. 𝑥 + 2 = 0, 𝑦 + 3 = 0
C. 𝑥 − 2 = 0, 𝑦 − 3 = 0
D. 𝑥 − 2 = 0, 𝑦 = 0

24. The asymptotes of the curve 𝑥 2 𝑦 2 − 𝑎2 (𝑥 2 + 𝑦 2 ) = 0 form a

A. circle
B. square
C. pentagon
D. triangle

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Notes
Calculus
Answer for Self Assessment
1. B 2. C 3. B 4. A 5. B

6. D 7. C 8. B 9. A 10. D

11. D 12. A 13. A 14. C 15. B

16. C 17. B 18. A 19. D 20. D

21. A 22. A 23. C 24. B

Review Questions
1. If the radius of circle A is 1/6, then its curvature is ____________.

B
A

2. If the radius of circle B is 1/2, then its curvature is ____________.

3. If the radius of circle C is 1/5, then its curvature is ____________.

4. If the radius of circle D is 1/3, then its curvature is ____________.

5. If you have a small circle and a large circle, which one will have the larger curvature?
6. Find the parallel asymptotes for the curve 2𝑥 3 − 𝑥 2 𝑦 + 2𝑥𝑦 2 + 𝑦 3 − 4𝑥 2 + 8𝑥𝑦 − 4𝑥 + 1 = 0.
7. Find the rectangular asymptotes for the curve 2𝑥 3 − 𝑥 2 𝑦 + 2𝑥𝑦 2 + 𝑦 3 − 4𝑥 2 + 8𝑥𝑦 = 0.
8.Find the rectangular asymptotes for the curve 𝑥 2 𝑦 − 3𝑥 2 − 5𝑥𝑦 + 6𝑦 + 2 = 0.
9. Find the parallel asymptotes for the curve 𝑥 2 𝑦 + 𝑥𝑦 2 + 𝑥𝑦 + 𝑦 2 + 3𝑥 = 0.
10. Find the oblique asymptotes for the curve 2𝑥 3 − 𝑥 2 𝑦 + 2𝑥𝑦 2 + 𝑦 3 − 4𝑥 2 + 8𝑥𝑦 − 4𝑥 + 1 = 0.
11. Find the oblique asymptotes for the curve 2𝑥 3 − 𝑥 2 𝑦 + 2𝑥𝑦 2 + 𝑦 3 − 4𝑥 2 + 8𝑥𝑦 = 0.
12. Find the oblique asymptotes for the curve 𝑥 2 𝑦 − 3𝑥 2 − 5𝑥𝑦 + 6𝑦 + 2 = 0.
13. Find the oblique asymptotes for the curve 𝑥 2 𝑦 + 𝑥𝑦 2 + 𝑥𝑦 + 𝑦 2 + 3𝑥 = 0.
14. Find all the asymptotes for the curve 𝑥 3 − 𝑥 2 𝑦 − 𝑥𝑦 2 + 𝑦 3 + 2𝑥 2 − 4𝑦 2 + 2𝑥𝑦 + 𝑥 + 𝑦 + 1 = 0.
15. Find all the asymptotes for the curve 𝑥 3 − 𝑥 2 𝑦 − 𝑥𝑦 2 + 𝑦 3 + 2𝑥 2 − 4𝑦 2 + 2𝑥𝑦 = 0.

Further Reading
W. Thomas Finny (1998). Calculus and Analytic Geometry, 6th Edition, Publishers, Narsa,
India.
R. K. Jain, and Iyengar, SRK. (2010). Advanced Engineering Mathematics, 3 rd Edition
Publishers, Narsa, India.
Widder, D.V. (2002). Advance Calculus 2nd Edition, Publishers, PHI, India.
Piskunov, N. (1996). Differential and Integral Calculus Vol I, & II, Publishers, CBS, India.

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 Notes
Unit 12: Curvature and Asymptotes

https://www.whitman.edu/mathematics/calculus_online/section13.03.html
https://www.math24.net/curvature-plane-curves
https://www.semanticscholar.org/paper/Asymptotes%2C-Cubic-Curves%2C-and-the-
Projective-Plane-Nunemacher/82e455119d0ad5d7ad8c56770792302ed4c89263?p2df
https://www.intmath.com/applications-differentiation/8-radius-curvature.php

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 Notes
Dr. Pratiksha, Lovely Professional University Unit 13: Concavity and Multiple Points

Unit 13: Concavity and Multiple Points

CONTENTS
Objectives
Introduction
13.1 Symmetry
13.2 Various Kinds of Symmetry
13.3 Concavity and Convexity
13.4 Points of Inflection
13.5 Tangents at Origin
13.6 Multiple Points
13.7 Condition for any Point (x, y) to be a Multiple Point of the Curve f(x, y)=0
13.8 Position and Nature of Double Points
Summary
Keywords
Self Assessment
Answer for Self Assessment
Review Questions
Further Reading

Objectives
Students will be able to

• detect the lines of symmetry in a curve

• classify a point as a point of concavity, convexity or inflection
• find the tangents at origin
• understand various multiple points
• find the position and nature of the double points

Introduction
In this unit, we will mainly learn about various important aspects involved in the tracing of a curve.
We will begin with the symmetry and its various aspects in relation to different shapes. From the
differentiability we have the idea of smoothness or pointedness of the curve at a point. This idea
can be extended to concavity of a function, with a special focus on the points of inflection. There
will be further a discussion on the types of double points, their nature and their position.

13.1 Symmetry
The images which can be divided into identical halves are called symmetrical. The images that
cannot be divided into identical halves are asymmetrical.

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Calculus

Any line splitting a shape into two parts such that the two parts are the same is called a line of
symmetry. These parts are also said to be symmetrical to each other.
For a square there are four lines of symmetry. for a hexagon, there will be six lines of symmetry.
Can you think of the lines of symmetry for a triangle and a pentagon?
You can take a piece of paper, draw the required shape on it, using the scale and pencil, cut out
them and fold it in various ways to find out the lines of symmetry.
Consider the folium of Descartes and the cardioid; you can observe that there is only one line of
symmetry.

13.2 Various Kinds of Symmetry

We can see that there are various lines of symmetry in various kinds
of functions 𝑓(𝑥, 𝑦) = 0. We can classify those lines as follows:

Symmetry about the x-axis:

In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑦 with – 𝑦. If (𝑥, −𝑦) = 𝑓(𝑥, 𝑦) ,
then the graph will be symmetric about the x-axis.
e.g. 𝑦 2 (2𝑎 − 𝑥) = 𝑥 3 is symmetric about the x-axis.

Symmetry about the y-axis:

In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with – 𝑥. If (−𝑥, 𝑦) = 𝑓(𝑥, 𝑦) , then the graph will be symmetric
about the y-axis.
e.g. (𝑥 2 + 𝑦 2 )2 = 𝑎2 (𝑥 2 − 𝑦 2 )is symmetric about the y-axis.

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 Notes
Unit 13: Concavity and Multiple Points
Symmetry about the origin:
In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥, 𝑦 with – 𝑥, −𝑦. If (−𝑥, −𝑦) = 𝑓(𝑥, 𝑦) , then the graph will be
symmetric about the origin.
e.g. (𝑥 2 + 𝑦 2 )2 = 𝑎2 (𝑥 2 − 𝑦 2 )is symmetric about the origin

Symmetry about the line,𝒚 = 𝒙:

In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with 𝑦 and 𝑦 with 𝑥.If (𝑥, 𝑦) = 𝑓(𝑦, 𝑥) , then the graph will be
symmetric about the line 𝑦 = 𝑥.
e.g. (𝑥 2 + 𝑦 2 )2 = 𝑎2 (𝑥 2 − 𝑦 2 )is not symmetric about the line 𝑦 = 𝑥.

Symmetry about the line, 𝒚 = −𝒙:

In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with −𝑦 and 𝑦 with −𝑥. If (𝑥, 𝑦) = 𝑓(−𝑦, −𝑥) , then the graph
will be symmetric about the line 𝑦 = −𝑥.
e.g. (𝑥 2 + 𝑦 2 )2 = 𝑎2 (𝑥 2 − 𝑦 2 )is not symmetric about the line 𝑦 = −𝑥.

Symmetry in the opposite quadrants:

In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥, 𝑦 with – 𝑥, −𝑦. If (−𝑥, −𝑦) = 𝑓(𝑥, 𝑦) , then the graph will be
symmetric in the opposite quadrants.
e.g. 𝑥𝑦 = 1 is symmetric in first and third quadrants.

Discuss all possible lines of symmetry for the following curves:

(i) 𝑦 2 (2𝑎 − 𝑥) = 𝑥 3
(ii) (𝑥 2 + 𝑦 2 )2 = 𝑎2 (𝑥 2 − 𝑦 2 )
(iii) (𝑥 − 𝑏)2 (𝑥 2 + 𝑦 2 ) − 𝑎2 𝑥 2 = 0
(iv) 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
(v) (𝑥 2 + 𝑦 2 − 𝑎𝑥𝑦)2 = 4𝑎2 (𝑥 2 + 𝑦 2 )
(vi) 𝑥 2 + 𝑦 2 = 16
(vii) 𝑦 2 = 2𝑥
(viii) 𝑥 2 + 𝑦 = 2𝑥 + 4

13.3 Concavity and Convexity

Although the first derivative test determines if a function is increasing or decreasing, we would
also like to know if the shape of the graph is curving upward or downward. This notion of
curvature of a graph upward or downward is known as concavity.
If the secant line passing through the points (x1, f(x1)) and (x2, f(x2)) is above the curve f(x)=y
between these two points, then f(x) is concave up.
If the secant line passing through the points (x 1, f(x1)) and (x2, f(x2)) is below the curve f(x)=y
between these two points, then f(x) is concave below or concave down.

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Calculus
When the slope continually increases, the function is concave upward.
When the slope continually decreases, the function is concave downward.

Theorem:
Consider a function f(x) that is twice continuously differentiable on an interval I. The
Function f(x) is

• concave upwards if f''(x) > 0 for all x in I

• concave downwards if f''(x) < 0 for all x in I

You need to be careful while using the following terms:

• Concave Downward is also called Concave or Convex Upward

• Concave Upward is also called Convex or Convex Downward

Discuss the curve 𝑦 = 𝑥 2 for its concavity.

𝑦 ′ (𝑥) = 2𝑥
Clearly the first derivative is positive whenever 𝑥 is positive and is negative, whenever 𝑥 is
negative. So by first derivative test, we can see that the slope is decreasing for negative values and
is increasing for positive values of 𝑥. Therefore the curve is concave up for all 𝑥.
Alternatively, we can find the second derivative
𝑦 ′′ (𝑥) = 2
Clearly, it is positive for all the values of the domain, thus the curve 𝑦 = 𝑥 2 is concave upwards
∀ 𝑥 ∈ 𝑹.

Discuss the curve 𝑓(𝑥) = 5𝑥 3 + 2𝑥 2 − 3𝑥 for its concavity.

𝑓 ′ (𝑥) = 15𝑥 2 + 4𝑥 − 3
𝑓 ′′ (𝑥) = 30𝑥 + 4
2
𝑓 ′′ (𝑥) = 0 happens at 𝑥 = −
15
2
Clearly 𝑓 ′′ (𝑥) < 0 in (−∞, − ) ⇒ 𝑓(𝑥) is concave downward in this interval and
15

2
𝑓 ′′ (𝑥) > 0 in (− , ∞) ⇒ 𝑓(𝑥) is concave upward in this interval.
15

Discuss the curve 𝑓(𝑥) = ln(1 + 𝑥 2 ) for its concavity.

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Unit 13: Concavity and Multiple Points
We have
2𝑥
𝑓 ′ (𝑥) =
1 + 𝑥2
2 − 2𝑥 2
𝑓 ′′ (𝑥) =
(1 + 𝑥 2 )2
We can look for the values of 𝑥 for which the 𝑓′′(𝑥) will be negative. That is
2 − 2𝑥 2 < 0
2(1 − 𝑥)(1 + 𝑥) < 0
From this inequality, it is clear that,𝑓(𝑥) is concave downward in (−∞, −1)𝑈(1, ∞) and concave
upward in (−1,1). Let us also look at how the function looks like!

13.4 Points of Inflection

A point where a curve changes from Concave upward to Concave downward (or vice versa), is
called the inflexion point.
A point (c, f(c)) is said to be an inflection point for a point c in (a, b) and for a continuous function
f(x) in (a, b) if the graph of y=f(x) changes concavity at (c, f(c)).
This also implies that the first derivative changes from increasing to decreasing or decreasing to
increasing at(c, f(c)).

Find the point of inflection for the curve 𝑓(𝑥) = 𝑥 𝑒 −2𝑥

Finding the derivatives, we get
𝑓 ′ (𝑥) = 𝑒 −2𝑥 (−2𝑥 + 1)
𝑓 ′′ (𝑥) = 𝑒 −2𝑥 (4𝑥 − 4)
The curve changes from concave upward to concave downward when
𝑓 ′′ (𝑥) = 0
And so we get 𝑥 = 1

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∴ the point of inflection is (1, 𝑒 −2 )

Find the point of inflection for the curve 𝑓(𝑥) = (𝑥 − 1)3 (𝑥 − 5)

Finding the derivatives, we get
𝑓 ′ (𝑥) = 4(𝑥 − 1)2 (𝑥 − 4)
𝑓 ′′ (𝑥) = 12(𝑥 − 1)(𝑥 − 3)
The curve changes from concave upward to concave downward when
𝑓 ′′ (𝑥) = 0
⇒ 𝑥 = 1, 3

Thus, the points of inflection are (1, 0) and (3, -16)

5
Find the point of inflection for the curve 𝑓(𝑥) = 𝑥 + 𝑥 3
Finding the derivatives, we get
5 2
𝑓 ′ (𝑥) = 1 + 𝑥 3
3
10
𝑓 ′′ (𝑥) = 1
9𝑥 3
Here the 𝑓 ′′ (𝑥) = 0 won’t make sense, so using the first derivative test, we can observe that
when𝑥 < 0, 𝑓 ′′ (𝑥) < 0 and when 𝑥 > 0, 𝑓 ′′ (𝑥) > 0.
So we can see the curve changes from concave downward to concave upward at (0, 0), So (0, 0) is
the point of inflection of the given curve.

Find the point of inflection for the curve 𝑓(𝑥) = 𝑥 4 − 6𝑥 2

Finding the derivatives, we get
𝑓 ′ (𝑥) = 4𝑥 3 − 12𝑥
𝑓 ′′ (𝑥) = 12𝑥 2 − 12
The curve changes from concave upward to concave downward when
𝑓 ′′ (𝑥) = 0
⇒ 𝑥 = ±1

The points of inflection are (1, -5) and (-1, -5).

13.5 Tangents at Origin

The general equation of rational algebraic curve of the 𝑛𝑡ℎ degree which passes through the origin
O, when arranged according to ascending powers of 𝑥 and 𝑦 is of the form
(𝑏1 𝑥 + 𝑏2 𝑦) + (𝑐1 𝑥 2 + 𝑐2 𝑥𝑦 + 𝑐3 𝑦 2 ) + (𝑑1 𝑥 3 + 𝑑2 𝑥 2 𝑦 + 𝑑3 𝑥𝑦 2 + 𝑑4 𝑦 3 ) + ⋯ = 0 ---- (1)
𝑦
Let 𝑃(𝑥, 𝑦) be any point on the curve. The slope of the chord OP is .
𝑥

lim 𝑐ℎ𝑜𝑟𝑑 𝑂𝑃 = Tangent at 𝑂

𝑃→𝑂

Then when 𝑥 → 0, 𝑦 → 0
𝑦
lim = 𝑚is the slope of the tangent.
𝑥

So (1) implies
𝑦 𝑦 𝑦
(𝑏1 + 𝑏2 ) + (𝑐1 𝑥 + 𝑐2 + 𝑐3 𝑦 ) + ⋯ = 0
𝑥 𝑥 𝑥
𝑏1 + 𝑏2 𝑚 = 0

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Unit 13: Concavity and Multiple Points
𝑏1
𝑚=−
𝑏2
𝑦 𝑏1
=−
𝑥 𝑏2
∴ 𝑏1 𝑥 + 𝑏2 𝑦 = 0 is the tangent at the origin, which can be written by equating the lowest degree
term to zero in equation (1).
If 𝑏2 = 0 but 𝑏1 ≠ 0, then considering the slope of OP with reference to y-axis, it can be shown that
the tangent retains the same form.
Let 𝑏1 = 𝑏2 = 0, then the equation takes the form
𝑐1 𝑥 2 + 𝑐2 𝑥𝑦 + 𝑐3 𝑦 2 + 𝑑1 𝑥 3 + ⋯ = 0
𝑥𝑦 𝑦2
𝑐1 + 𝑐2 + 𝑐3 + 𝑑1 𝑥 + ⋯ = 0
𝑥2 𝑥2
As 𝑥 → 0
𝑐1 + 𝑐2 𝑚 + 𝑐3 𝑚2 = 0 ---- (2)
gives two values of 𝑚 say 𝑚1 and 𝑚2 .
The equation of either tangent at the origin is 𝑦 = 𝑚1 𝑥 ---- (3)
Eliminating 𝑚1 from (2) and (3), we get
𝑐1 𝑥 2 + 𝑐2 𝑥𝑦 + 𝑐3 𝑦 2 = 0 ----(4)
whichis the joint equation of two tangents at the origin, and it can also be written by equating the
lowest degree term to zero in the equation of the curve.
If 𝑐1 = 𝑐2 = 𝑐3 = 0, then (4) becomes an identity and equations of tangents can still be written by
equating to zero, the terms of lowest degree, which is third, in this case.
Therefore the rule to find the tangents at origin can be summarized as:

The equation of the tangent or tangents at the origin is obtained by equating to zero
the terms of the lowest degree in the equation of the curve .

Find the tangents at origin for the curve 𝑥 3 + 𝑦 3 − 3𝑎𝑥𝑦 = 0

Clearly the origin lies on the curve. To find the tangents at origin, let us seek the lowest degree term
in the curve and equate that to zero.
−3𝑎𝑥𝑦 = 0
⇒ 𝑥 = 0, 𝑦 = 0

are the required equations of the tangents at the origin.

Find the equation of the tangent(s) at (-1,-2) to the curve 𝑥 3 + 2𝑥 2 + 2𝑥𝑦 − 𝑦 2 + 5𝑥 − 2𝑦 = 0.

We can shift the origin to the point (-1, -2) by the following transformations:
𝑥 =𝑋−1
𝑦=𝑌−2
Using these, we get the transformed equation as:
𝑋 3 − 𝑋 2 + 2𝑋𝑌 − 𝑌 2 = 0
Clearly the origin lies on the curve and we can find the tangents at origin easily by equating to zero
the terms of the lowest degree in the equation of the curve .
i.e.– 𝑋 2 + 2𝑋𝑌 − 𝑌 2 = 0
i.e.(𝑋 − 𝑌)2 = 0
i.e.𝑋 − 𝑌 = 0
i.e.𝑥 + 1 − (𝑦 + 2) = 0

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Calculus
i.e. 𝑥 − 𝑦 = 1 is the tangent at (-1, -2) for the given curve.
Therefore, we can find the tangent at any given point on a curve.

13.6 Multiple Points

If RS is an arc of a curve, and if at the point P on RS there exists one and only one tangent, AB, to
the curve, then point P is known as an ordinary point of the curve.
If at a point P on a curve there exist two and only two distinct tangents, then that point is called a
node.
If the two tangents at a given point are not distinct, but coincide, we have what is called a cusp.

There are several kinds of cusps:

(1) If the curve in the neighbourhood of a cusp lies partly on one side of the tangent and partly on
the other side, the point is known as a cusp of the first kind;
(2) If the curve lies entirely on one side of the common tangent (in the region of tangency), the point
is known as a cusp of the second kind;
(3) If there are two distinct cusps at the same point, it is known as a point of osculation.
All points having two and only two tangents, whether real or imaginary, distinct or coincident, are
called double points of the curve.
Thus nodes and cusps of all kinds are double points.
Triple points are such points on a curve for which there are three tangents; similarly there are four
for quadruple points etc.
An isolated point on a curve is also called a conjugate point.
All points that are not ordinary points are known as singular points.

13.7 Condition for any Point (x, y) to be a Multiple Point of the Curve
f(x, y)=0
For a curve 𝑓(𝑥, 𝑦) = 0 ---- (1)
We can write

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Unit 13: Concavity and Multiple Points
𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦
+ =0
𝜕𝑥 𝑑𝑥 𝜕𝑦 𝑑𝑥
𝑑𝑦
𝑓𝑥 + 𝑓𝑦 =0
𝑑𝑥
𝑑𝑦
At a multiple point of a curve, the curve has at least two tangents and must have at least two
𝑑𝑥
values at multiple points.
𝑑𝑦 𝑑𝑦
More than one value of can satisfy 𝑓𝑥 + 𝑓𝑦 = 0iff 𝑓𝑥 = 0, 𝑓𝑦 = 0.
𝑑𝑥 𝑑𝑥

Therefore, to find the multiple points, we have to find the values of (𝑥, 𝑦) which simultaneously
satisfy the three equations:
𝑓𝑥 (𝑥, 𝑦) = 0
𝑓𝑦 (𝑥, 𝑦) = 0
𝑓(𝑥, 𝑦) = 0
𝑑𝑦
Differentiate 𝑓𝑥 + 𝑓𝑦 = 0 w.r.t. 𝑥, we get
𝑑𝑥

𝑑𝑦 𝑑2 𝑦 𝑑𝑦 2
𝑓𝑥𝑥 + 2𝑓𝑥𝑦 + 𝑓𝑦 2 + 𝑓𝑦𝑦 ( ) = 0
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦
At the multiple points where 𝑓𝑦 = 0, 𝑓𝑥 = 0; the value of are the roots of the quadratic equation,
𝑑𝑥

𝑑𝑦 𝑑𝑦 2
𝑓𝑥𝑥 + 2𝑓𝑥𝑦 + 𝑓𝑦𝑦 ( ) = 0
𝑑𝑥 𝑑𝑥
In case 𝑓𝑥𝑥 , 𝑓𝑥𝑦 , 𝑓𝑦𝑦 are not all zero and 𝑓𝑥 = 0 = 𝑓𝑦 , the point (𝑥, 𝑦) will be a double point and will
be a
node if 𝒇𝟐𝒙𝒚 − 𝒇𝒙𝒙 𝒇𝒚𝒚 > 0

cusp if 𝒇𝟐𝒙𝒚 − 𝒇𝒙𝒙 𝒇𝒚𝒚 = 𝟎 and

conjugate point if 𝒇𝟐𝒙𝒚 − 𝒇𝒙𝒙 𝒇𝒚𝒚 < 0

If 𝑓𝑥𝑥 = 𝑓𝑥𝑦 = 𝑓𝑦𝑦 = 0, the point (𝑥, 𝑦) will be a multiple point of the order higher than two,
those are not in the scope of this course.

Find the multiple points on the curve

𝑥 4 − 2𝑎𝑦 3 − 3𝑎2 𝑦 2 − 2𝑎2 𝑥 2 + 𝑎4 = 0
Let 𝑓(𝑥) = 𝑥 4 − 2𝑎𝑦 3 − 3𝑎2 𝑦 2 − 2𝑎2 𝑥 2 + 𝑎4
𝑓𝑥 = 4𝑥 3 − 4𝑎2 𝑥
𝑓𝑦 = −6𝑎𝑦 2 − 6𝑎2 𝑦
𝑓𝑥 = 0, 𝑓𝑦 = 0 ⇒ 𝑥 = 0, 𝑎, −𝑎 ; 𝑦 = 0, −𝑎.

∴ 𝑓𝑥 and𝑓𝑦 vanish at (0,0), (0, −𝑎), (𝑎, 0), (𝑎, −𝑎), (−𝑎, 0), (−𝑎, −𝑎).
Out of these, only (𝑎, 0), (−𝑎, 0)& (0, −𝑎) lie on the given curve.
Now 𝑓𝑥𝑥 = 12𝑥 2 − 4𝑎2
𝑓𝑦𝑦 = −12𝑎𝑦 − 6𝑎2
𝑓𝑥𝑦 = 0
We know,

𝑑𝑦 𝑑𝑦 2
𝑓𝑥𝑥 + 2𝑓𝑥𝑦 + 𝑓𝑦𝑦 ( ) = 0
𝑑𝑥 𝑑𝑥
At (𝑎, 0)

𝑑𝑦 2
−6𝑎2 ( ) + 4(2𝑎2 ) = 0
𝑑𝑥

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𝑑𝑦 2
=±
𝑑𝑥 √3
Since there are two real values, the point (𝑎, 0) is a node and the tangents at (𝑎, 0) are
2
𝑦=± (𝑥 − 𝑎)
√3
2
Similarly tangents at (−𝑎, 0): 𝑦 = ± (𝑥 + 𝑎)
√3

√2
and at (0, −𝑎): 𝑦 + 𝑎 = ± 𝑥
√3

Alternatively,
We can find tangents at (𝑎, 0), by shifting the origin to (𝑎, 0) by following transformations,
𝑥 = 𝑋 + 𝑎, 𝑦 = 𝑌
⇒ 𝑋 + 4𝑋 𝑎 − 2𝑎𝑌 3 + 4𝑎2 𝑋 2 − 3𝑎2 𝑌 2 = 0
4 3

Tangents at origin are given by,

4𝑎2 𝑋 2 − 3𝑎2 𝑌 2 = 0
3
𝑋 2 = 𝑌2
4
3
𝑋=± 𝑌
√2
Similarly the tangents can be found at the other points too.

13.8 Position and Nature of Double Points

For the curve 𝑓(𝑥, 𝑦) = 0, (𝑥, 𝑦) is a double point if it satisfies
𝑓𝑥 (𝑥, 𝑦) = 0
𝑓𝑦 (𝑥, 𝑦) = 0
𝑓(𝑥, 𝑦) = 0
Simultaneously.
The point (𝑥, 𝑦) is a node, if 𝑓𝑥𝑦
2 −𝑓 𝑓 >0
𝑥𝑥 𝑦𝑦

The point (𝑥, 𝑦) is a cusp, if 𝑓𝑥𝑦

2
− 𝑓𝑥𝑥 𝑓𝑦𝑦 = 0
The point (𝑥, 𝑦) is isolated, if 𝑓𝑥𝑦
2 −𝑓 𝑓 <0
𝑥𝑥 𝑦𝑦

Types of cusps in terms of their position:

Single cusp of first kind: Two branches on the same side of common normal and on opposite
sides of tangent.
Single cusp of second kind: Two branches lie on the same side of the normal as well as the
tangent.
Double cusp of second kind: Two branches lie on the different sides of normal and on the same
side of the tangent.
Point of oscu inflection: Two branches lie on different sides of normal but on one side they lie on
the same and on the other on opposite sides of the common tangent.
Let us learn about the position and nature of double points by an example.

Locate the double points of the curve 𝑦(𝑦 − 6) = 𝑥 2 (𝑥 − 2)3 − 9 and mention their nature.
Here 𝑓(𝑥, 𝑦) = 𝑥 2 (𝑥 − 2)3 − 9 − 𝑦(𝑦 − 6) = 0
𝑓𝑥 = 𝑥(𝑥 − 2)2 (5𝑥 − 4)
𝑓𝑦 = 6 − 2𝑦

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Unit 13: Concavity and Multiple Points
𝑓𝑥𝑥 = (𝑥 − 2)2 (5𝑥 − 4) + 2𝑥(𝑥 − 2)(5𝑥 − 4) + 5𝑥(𝑥 − 2)2
𝑓𝑦𝑦 = −2
𝑓𝑥𝑦 = 0
4
Now 𝑓𝑥 = 0, 𝑓𝑦 = 0 imply 𝑥 = 0, 2, and 𝑦 = 3
5

Therefore possible double points are

4
(0,3), (2, 3)and ( , 3)
5
4
The point ( , 3) does not satisfy the given curve. So there are only two double points.
5

At (0, 3)
2 −𝑓 𝑓 < 0
𝑓𝑥𝑦 𝑥𝑥 𝑦𝑦

So, (0, 3) is a conjugate point.

At (2, 3)
2 −𝑓 𝑓 = 0
𝑓𝑥𝑦 𝑥𝑥 𝑦𝑦

So, (2, 3) is a cusp.

To know the nature of the cusp, let us shift the origin to the point (2, 3) by using
𝑥 =𝑋+2
𝑦=𝑌+3
The transformed equation is 𝑌2 = 𝑋 3 (𝑋 + 2)2
The tangents at origin are 𝑌2 = 0 i.e. the x-axis.

Moreover, 𝑌 = ±(𝑋 + 2)√𝑋 3

When 𝑋 < 0, 𝑌 is imaginary
When 𝑋 > 0, 𝑌has two values, positive and negative. Therefore near the origin, the curve lies on
both sides of X-axis (tangent) and only on one side of the Y-axis (normal). The graph is given below
for better clarity.

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Therefore, the new origin (2,3) is a single cusp of first kind.

Summary
This chapter had variety of topics, which can be summarized as follows:

• Symmetry about the x-axis: In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑦 with – 𝑦. If (𝑥, −𝑦) =
𝑓(𝑥, 𝑦) , then the graph will be symmetric about the x-axis.
• Symmetry about the y-axis: In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with – 𝑥. If (−𝑥, 𝑦) =
𝑓(𝑥, 𝑦) , then the graph will be symmetric about the y-axis.
• Symmetry about the origin: In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥, 𝑦 with – 𝑥, −𝑦. If
(−𝑥, −𝑦) = 𝑓(𝑥, 𝑦) , then the graph will be symmetric about the origin.
• Symmetry about the line, 𝑦 = 𝑥:In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with 𝑦 and 𝑦 with 𝑥.
If (𝑥, 𝑦) = 𝑓(𝑦, 𝑥) , then the graph will be symmetric about the line 𝑦 = 𝑥.
• Symmetry about the line, 𝑦 = −𝑥:In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥 with −𝑦 and 𝑦 with
−𝑥. If (𝑥, 𝑦) = 𝑓(−𝑦, −𝑥) , then the graph will be symmetric about the line 𝑦 = −𝑥.
• Symmetry in the opposite quadrants: In the function 𝑓(𝑥, 𝑦) = 0, replace 𝑥, 𝑦 with – 𝑥, −𝑦.
If (−𝑥, −𝑦) = 𝑓(𝑥, 𝑦) , then the graph will be symmetric in the opposite quadrants.
• Consider a function f(x) that is twice continuously differentiable on an interval I. The
function f(x) is concave upwards if f''(x) > 0 for all x in I and concave downwards if f''(x) <
0 for all x in I
• A point where a curve changes from concave upward to concave downward (or vice
versa), is called the inflexion point.

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Unit 13: Concavity and Multiple Points
• The equation of the tangent or tangents at the origin is obtained by equating to zero the
terms of the lowest degree in the equation of the curve .
• The point (𝑥, 𝑦) will be a double point and will be a node if 𝑓𝑥𝑦
2 − 𝑓 𝑓 > 0, cusp if 𝑓 2 −
𝑥𝑥 𝑦𝑦 𝑥𝑦

𝑓𝑥𝑥 𝑓𝑦𝑦 = 0 and conjugate point if 𝑓𝑥𝑦

2 −𝑓 𝑓 < 0
𝑥𝑥 𝑦𝑦

Keywords
Symmetry, lines of symmetry, concave up, concave down, convex, concave, point of inflection,
tangent at origin, multiple point, double point, node, cusp, isolated point

Self Assessment
1. How many lines of symmetry are there for a pentagon?
A. 2
B. 3
C. 4
D. 5

2. For a lemniscate of Bernoulii, how many lines of symmetry are there?

A. 2
B. 3
C. 4
D. 5

3. The curve 2𝑥 2 + 2𝑦 2 = 11 is
A. symmetric about the x-axis
B. symmetric about the y-axis
C. symmetric about the line y=x
D. all of the above
4. The notion of curvature of a graph upward or downward is known as
A. symmetry
B. asymptotes
C. concavity
D. multiple points

5. When the slope continually increases, the function

A. is concave upwards
B. is concave downwards
C. can not be deciphered
D. is decreasing

6. The function 𝑓(𝑥) = 2𝑥 2 + 3𝑥 + 4 is

A. concave upwards on the set of real numbers
B. concave upwards on a specific interval
C. concave downwards on the set of real numbers
D. concave downwards on a specific interval

7. The point where a curve changes from concave upward to concave downward is called a
A. saddle point

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Calculus
B. stationary point
C. critical point
D. inflexion point

8. Which of the following is true for the point of inflexion ( c, f(c) )?

A. the first derivative changes from increasing to decreasing or decreasing to increasing at (c,
f(c))
B. the first derivative changes from increasing to decreasing at (c, f(c))
C. the first derivative changes from decreasing to increasing at (c, f(c))
D. none of these

9. For 𝑓(𝑥) = 𝑥𝑒 −2𝑥 the point of inflexion is

A. (1,1)
B. (1,e)
C. (1, 𝑒 −2 )
D. (1, 𝑒 2 )

10. The equation of the tangent or tangents at the origin is obtained by

A. equating to zero the terms of the lowest degree in the equation of the curve.
B. equating to zero the terms of the highest degree in the equation of the curve.
C. equating to one the terms of the lowest degree in the equation of the curve.
D. equating to one the terms of the highest degree in the equation of the curve.

11. For the curve 𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0, there exist ____ tangent(s) at the origin.
A. one
B. two
C. three
D. four

12. For the curve 𝑥 2 (𝑥 2 + 𝑦 2 ) = 5(𝑥 − 𝑦), the equation of the tangent at origin is
A. 𝑥 = 5𝑦
B. 𝑥 = −5𝑦
C. 𝑥 = −𝑦
D. 𝑥 = 𝑦

13. If the two tangents at a given point are not distinct, but coincide, we get
A. a node
B. a cusp
C. a conjugate point
D. none of these

14. If the curve lies entirely on one side of the common tangent (in the region of tangency),
the point is known as a
A. cusp of first kind
B. cusp of second kind

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Unit 13: Concavity and Multiple Points
C. node
D. point of osculation

15. At a multiple point of a curve 𝑓(𝑥) = 0, the curve has

A. one tangent
B. two tangents
C. at least one tangent
D. at least two tangents

16. If at a point on a curve there exist two and only two distinct tangents, then that point is
called a
A. cusp of first kind
B. cusp of second kind
C. node
D. point of osculation

2 − 𝑓 𝑓 = 0, then the double point is a

17. At a double point (𝑥, 𝑦) of a curve 𝑓(𝑥, 𝑦) = 0, if 𝑓𝑥𝑦 𝑥𝑥 𝑦𝑦

A. node
B. cusp
C. isolated point
D. none of these

2 − 𝑓 𝑓 > 0, then the double point is a

18. At a double point (𝑥, 𝑦) of a curve 𝑓(𝑥, 𝑦) = 0, if 𝑓𝑥𝑦 𝑥𝑥 𝑦𝑦

A. node
B. cusp
C. isolated point
D. none of these

19. At a double point (𝑥, 𝑦) of a curve 𝑓(𝑥, 𝑦) = 0, if 𝑓𝑥𝑦

2 − 𝑓 𝑓 < 0, then the double point is a
𝑥𝑥 𝑦𝑦

A. node
B. cusp
C. isolated point
D. none of these

Answer for Self Assessment

1. D 2. A 3. D 4. C 5. A

6. A 7. D 8. A 9. C 10. A

11. B 12. D 13. B 14. B 15. D

16. C 17. B 18. A 19. C

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Calculus
Review Questions
1. Draw any random closed figure with 8 straight lines and discuss its lines of symmetry.
2. In the nature around, spot five things, having symmetry.
3. Discuss all the lines of symmetry for the curve (𝑥 2 + 𝑦 2 )𝑥 − 2𝑎𝑦 2 = 0
4. Discuss about the concavity of the curve 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
5. Discuss about the points of inflection of the curve 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
6. Find the tangents at origin for the curve (𝑥 2 + 𝑦 2 )𝑥 − 2𝑎𝑦 2 = 0
7. Find the tangents at origin for the curve (𝑥 2 + 𝑦 2 )2 = 𝑎2 𝑥 2 + 𝑏 2 𝑦 2
8. Find the tangents at origin for the curve 2𝑦 5 + 5𝑥 5 − 3𝑥(𝑥 2 − 𝑦 2 ) = 0
9. Find the tangents at origin for the curve (𝑥 2 + 𝑦 2 )𝑥 2 = 𝑎(𝑥 − 𝑦)
10. Find the double points of the curve 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
11. Find the double points of the curve (𝑥 2 + 𝑦 2 )𝑥 − 2𝑎𝑦 2 = 0
12. Find the position and nature of the double point of the curve 𝑥 2 𝑦 2 = (𝑎 + 𝑦)2 (𝑏 2 − 𝑦 2 ) for
𝑎 < 𝑏and𝑎 > 𝑏.

Further Reading
W. Thomas Finny (1998). Calculus and Analytic Geometry, 6th Edition, Publishers, Narsa,
India.
R. K. Jain, and Iyengar, SRK. (2010). Advanced Engineering Mathematics, 3 rd Edition
Publishers, Narsa, India.
Widder, D.V. (2002). Advance Calculus 2nd Edition, Publishers, PHI, India.
Piskunov, N. (1996). Differential and Integral Calculus Vol I, & II, Publishers, CBS, India.

https://www.cuemath.com/geometry/symmetry/
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/03%3A_
The_Graphical_Behavior_of_Functions/3.04%3A_Concavity_and_the_Second_Derivative
https://schoolbag.info/mathematics/calculus_2/39.html
https://easy-to-understand-maths.blogspot.com/2019/03/multiple-point.html

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Dr. Pratiksha, Lovely Professional University Unit 14: Tracing of Curves

Unit 14: Tracing of Curves

CONTENTS
Objectives
Introduction
14.1 Procedure for Tracing Curves Given in Cartesian Equations
14.2 Tracing of Polar Curves
14.3 Tracing of Parametric Curves
Summary
Keywords
Self Assessment
Answers for Self Assessment
Review Questions
Further Reading

Objectives
Students will be able to

• list the properties which can be used to trace a curve

• trace the Cartesian curves
• trace the parametric curves
• trace the polar curves

Introduction
Can you recognize which of the following options depicts the curve x2y2(x+ y) = 1?

In order to answer the above question, we need to analyze the given Cartesian equation
thoroughly. An image is worth a thousand words. A curve which is the visual synonym of a
functionalrelation gives us the whole idea of information about the relation. Of course, we can also
obtain this information by analysing the equation which defines the functional relation. But
studying the associated curve is often easier and quicker. In addition to this, a curve which
represents a relation between two quantities also helps us to easily find the value of one quantity

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corresponding to a specific value of the other. In this unit we shall be using many results from the
previous units and will try to understand what is meant by the graph of a relation like f(x, y) = 0,
and how one can draw it.

Recall that, that the set of points {(x, y) : f(x, y) = 0) is known as the graph of the functional relation
f(x, y) = 0. Graphing a function or a functional relation means showing the points of the
corresponding set in a plane, thus, essentially curve tracing means plotting the points which satisfy
a given relation. However, there are some difficulties involved in this. Let's see what these are and
how to overcome them.
It is often not possible to plot all the points on a curve. The standard technique is to plot some
suitable points and to get a general idea of the shape of the curve by considering tangents,
asymptotes, singular points, extreme points, inflection points, concavity, monotonicity, periodicity
etc. Then we draw a free hand curve as nearly satisfying the various properties as is possible.
The curves or graphs that we draw have a limitation. If the range of values of either (or both)
variable is not finite, then it is not possible to draw the complete graph. In such cases the graph is
not only approximate, but is also incomplete. For example, consider the simplest curve, a straight
line. Suppose we want to draw the graph of 𝑓: 𝑅 → 𝑅such that𝑓(𝑥) = 𝑐. We know that this is in line
parallel to the x-axis. But it is not possible to draw a complete graph as the line extends infinitely on
both sides.
Suppose the equation of a curve is f(x, y) = 0.We shall now list some steps which, when taken, will
simplify our job of tracing this curve.

14.1 Procedure for Tracing Curves Given in Cartesian Equations

Steps for curve sketching (preferably in the same order) are summarized below:

1. Domain
The first step is to determine the extentof the curve. In other words we try to find a region or
regions of the plane which cannot have any point of the curve. For example, no point on the curve
𝑥 = 𝑦 2 , lies in the second or the third quadrant, as the x-coordinate of any point on the curve has to
be non-negative. This means that our curve lies entirely in the first and the fourth quadrants.
Note thatit is easier to determine the extent of a curve if its equation can be written explicitly as y =
f(x) or x =f(y).

2. Intercepts
The next step is to determine the points where the curve intersects the axes. If we put y = 0 in f(x, y)
= 0, and solve the resulting equation for x, we get the points of intersection with the x-axis.
Similarly, putting x = 0 and solving the resulting equation for y, we can find the points of
intersection with the y-axis.

3. Symmetry
We find out if the curve is symmetrical about any line, or about the origin. We have already
discussed symmetry of curves in the previous unit. This step reduces our workload. If the curve is
symmetric about the x-axis, we can focus upon the region above x-axis only and then can replicate
that for the complete curve.

4. Asymptotes
The next step is to find the asymptotes, if there are any. They indicate the trend of the branches of
the curve extending to infinity. Asymptotes, if they exist, provide a frame for the curve.

5. Intervals of increase and decrease

Calculate dy/dx. This will help you in locating the portions where the curve is rising (dy/dx >0) or
falling (dy/dx <0) or the points where it has a corner (dy/dx does not exist).

6. Local maximum and minimum

𝑑2𝑦
Calculate𝑑2 𝑦/𝑑𝑥 2 . This will help you in locating maxima (dy/dx = 0, < 0) and minima (dy/dx =
𝑑𝑥 2
𝑑2𝑦
0, >0).
𝑑𝑥 2

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Unit 14: Tracing of Curves
7. Concavity/Convexity/Points of inflection/Multiple points
𝑑2𝑦
Calculate𝑑2 𝑦/𝑑𝑥 2 . This will help you in locating maxima (dy/dx = 0, < 0) and minima (dy/dx =
𝑑𝑥 2
𝑑2𝑦
0, >0) along with concave up or concave down nature of the curve. You will also be able to
𝑑𝑥 2
𝑑2𝑦
determine the points of inflection ( 2 = 0 ). These will give you a good idea about the shape of the
𝑑𝑥
curve.
Another important step is to determine the singular points. The shape of the curve at these points
is, generally, more complex, as more than one branch of the curve passes through them. Find out
whether the origin lies on the curve. If it does, then find the equations of the tangents at the origin
by equating to zero the lowest degree terms and we can look out for cusps and nodes.
8. Graph of the Function
All the information obtained from above steps, finally has to be put on the x-y plane. Plot as many
points as you can, around the points already plotted. Also try to draw tangents to the curve at some
of these plotted points. For this you will have to calculate the derivative as these points. Now join
the plotted points by a smooth curve (except at points of discontinuity). The tangents will guide
you in this, as they give you the direction of the curve and the graph has to be traced then.
Warm up by tracing the simple popular curves like modulus function, exponential function,
parabola etc. using the above mentioned steps. Now let us trace the curve x2y2(x+ y) = 1 which
was asked in the introductory section.
Since the function is not an explicit function of x or y, we can skip the domain part.
When x=0, we cannot find y and when y=0, we cannot find x. It means that the curve does not
make any intercepts on the axes.
The curve is symmetric about the line y=x only.
There are no horizontal and vertical asymptotes, but an oblique asymptote y=-x.
And by these many steps only, the correct answer out of four options can be inferred.
If not we would have undertaken next steps too.

Trace 𝑦 = 𝑥 3 − 12𝑥 − 16
Let us trace the curve step by step. For your ease the steps are mentioned below:

1. Domain
The function can take all real values of 𝑥 as its domain.
2. Intercepts
When 𝑥 = 0, 𝑦 = −16
When 𝑦 = 0, 𝑥 = −2, 4
Therefore the given curve meets the axes at (0, −16), (−2,0), (4,0).
3. Symmetry
No particular line of symmetry exists.
4. Asymptotes
No asymptote exists
5. Intervals of increase and decrease
𝑦 ′ = 3𝑥 2 − 12
𝑦 ′ = 0 ⇒ 𝑥 = ±2
∴ 𝑦 is increasing in (−∞, −2)
𝑦 is decreasing in (−2, 2)
𝑦 is increasing in (2, ∞)
6. Local maximum and minimum
𝑦 ′′ = 6𝑥
At 𝑥 = −2, 𝑦 ′′ < 0 ⇒ (−2, 0)is a point of maximum.

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At 𝑥 = 2, 𝑦 ′′ > 0 ⇒ (2, −32)is a point of minimum.
7. Concavity/Convexity/Points of Inflection/Multiple points
𝑦 ′′ < 0 when 𝑥 < 0 ⇒ Concave downward
𝑦 ′′ > 0 when 𝑥 > 0 ⇒ Concave upward
8. Graph of the Function

14.2 Tracing of Polar Curves

A polar equation is any equation that describes a relation between r and θ, where r represents the
distance from the pole to a point on a curve, and θ represents the anti-clockwise angle made by a
point on a curve, the pole, and the initial line.

One advantage of using polar equations is that certain relations that are not functions in Cartesian
form can be expressed as functions in polar form.
Another advantage is that certain relations are much simpler to express in polar form rather than
Cartesian form.

Steps to trace a polar curve:

1. Symmetry
Consider a curve generated by the function r=f(θ) in polar coordinates.

• The curve is symmetric about the polar axis if for every point (r, θ) on the graph, the point
(r,−θ) is also on the graph. Similarly, the equation r=f(θ) is unchanged by replacing θ with
−θ
• The curve is symmetric about the pole if for every point (r, θ) on the graph, the point
(r,π+θ) is also on the graph. Similarly, the
equation r=f(θ) is unchanged when
replacing r with −r, or θ with π+θ.
• The curve is symmetric about the vertical
line θ=π/2 if for every point (r, θ) on the
graph, the point (r,π−θ) is also on the
graph. Similarly, the equation r=f(θ) is
unchanged when θ is replaced by π−θ.
2. Extent

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Unit 14: Tracing of Curves
(i) Find the limits within which r must lie for the permissible values of 𝜃. If r < a (r >a) for some a >
0, then the curve lies entirely within (outside) the circle r = a. (ii) If 𝑟 2 is negative for some values of
𝜃, then the curve has no portion in the corresponding region.
3. Pole
The curve passes through the pole, if for r = 0, there corresponds a real value of θ
4. Asymptote
If O is the pole and 𝑃(𝑟, 𝜃) is any point in the polar system PY is the line which is the asymptote,
and 𝑂𝑌 ⊥ 𝑃𝑌 .The polar equation of any line is 𝑝 = 𝑟 cos(𝜃 − 𝛼) where 𝑝 is the length of
perpendicular from the pole to the line and 𝛼 is the angle which this perpendicular makes with the
initial line.
𝑂𝑌 = 𝑝
∠𝑋𝑂𝑌 = 𝛼
If 𝑃(𝑟, 𝜃) be any point on the line then ∠𝑃𝑂𝑌 = 𝜃 − 𝛼
𝑝
and = cos(𝜃 − 𝛼)
𝑟

∴ 𝑝 = 𝑟 cos(𝜃 − 𝛼)is the required equation of the line which is the prospective asymptote.
5. Region
Find the region in which the curve does not exist, or find the greatest and least numerical value of r
etc.
6. Specific points
Trace the variation of r as θ varies.

Trace 𝑟 = 𝑎(1 + 𝑐𝑜𝑠 𝜃)

1. If we replace 𝜃to– 𝜃, the equation remains unchanged. So the given curve is symmetric
about the initial line.
2. The extent can be seen from the fact that maximum value of cos 𝜃is 1 and minimum value
is -1. Thus 𝑟 can take values from 0 to 2𝑎.
3. For 𝑟 = 0, we can see that𝜃 = 𝜋. This means that the curve passes through the pole.
4. Since it’s a polar curve we can find the value of 𝑟 at various 𝜃 values and then plot the
curve on a polar plane.
θ 0 𝜋 𝜋 𝜋 𝜋 2𝜋 𝜋
6 4 3 2 3
2𝑎 (√3 + 2) (√2 + 1) 3 𝑎 𝑎 0
𝑟 𝑎 𝑎 𝑎 2
2 √2 2

From all the above points, the following curve can be traced:

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14.3 Tracing of Parametric Curves

When the path of a particle moving in the plane is not the graph of a function, we cannot describe it
using a formula that expresses y directly in terms of x, or x directly in terms of y. Instead, we need
to use a third variable t, called a parameter and write x = f(t) y = g(t).
The set of points (x, y) = (f(t), g(t)) described by these equations when t varies in an interval I form a
curve, called a parametric curve, and x = f(t), y = g(t) are called the parametric equations of the
curve.

Trace the curve with parametric equations:

𝑥 = 𝑎 cos3 𝜃
𝑦 = 𝑏 sin3 𝜃
First of all we can make a table of variation of 𝑥 and 𝑦with 𝜃.

𝜃 0 𝜋 𝜋 3𝜋 2𝜋
2 2

𝑥 𝑎 0 −𝑎 0 𝑎

𝑦 0 𝑏 0 −𝑏 0

Now the (𝑥, 𝑦)points can be plotted on the 𝑥𝑦 −plane.

For the parametric equations
𝑑𝑦
𝑑𝑦 𝑑𝜃 𝑑𝑥
= 𝑑𝑥
, ≠0
𝑑𝑥 𝑑𝜃
𝑑𝜃
𝑑𝑦 𝑏 𝑑𝑥
Therefore = − tan 𝜃 provided that ≠0
𝑑𝑥 𝑎 𝑑𝜃
𝑑𝑥 𝜋 3𝜋
Moreover = 0 for 𝜃 = 0, , 𝜋, , 2𝜋
𝑑𝜃 2 2
𝑑𝑦
When 𝜃 = 0, =0
𝑑𝑥

At (𝑎, 0), the tangent is given as,

𝑑𝑦
𝑦−0 = ( ) (𝑥 − 𝑎)
𝑑𝑥 𝜃=0
⇒ 𝑦=0
𝜋 𝑑𝑦
When 𝜃 = , =∞
2 𝑑𝑥

At (0, 𝑏), the tangent is given as,

𝑑𝑦
𝑦 − 𝑏 = ( ) 𝜋 (𝑥 − 0)
𝑑𝑥 𝜃=
2

⇒ 𝑥=0
𝑑𝑦
When 𝜃 = 𝜋, =0
𝑑𝑥

At (−𝑎, 0), the tangent is given as,

𝑑𝑦
𝑦−0=( ) (𝑥 + 𝑎)
𝑑𝑥 𝜃=𝜋
⇒ 𝑦=0
3𝜋 𝑑𝑦
When 𝜃 = , =∞
2 𝑑𝑥

At (0, −𝑏), the tangent is given as,

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Unit 14: Tracing of Curves
𝑑𝑦
𝑦+𝑏 =( ) (𝑥 − 0)
𝑑𝑥 𝜃=3𝜋
2

⇒ 𝑥=0

Thus, the tangent at points (𝑎, 0)and (−𝑎, 0) is the x-axis and at the points (0, 𝑏)and (0, −𝑏) is the y-
axis.
And from this information we can draw the following:

Trace the curve with parametric equations:

𝑥 = 𝑎(𝜃 + 𝑠𝑖𝑛𝜃)
𝑦 = 𝑎(1 + 𝑐𝑜𝑠𝜃)
First of all we can make a table of variation of 𝑥 and 𝑦with𝜃.

𝜃 0 𝜋 −𝜋 3𝜋 −3𝜋

𝑥 0 𝑎𝜋 − 𝑎𝜋 3𝜋𝑎 −3𝜋𝑎

𝑦 2𝑎 0 0 0 0

𝑑𝑦
𝑑𝑦 𝑑𝜃 𝑑𝑥
= 𝑑𝑥
, ≠0
𝑑𝑥 𝑑𝜃
𝑑𝜃
𝑑𝑥
Here = 𝑎(1 + 𝑐𝑜𝑠𝜃)
𝑑𝜃

𝑑𝑦
= −𝑎 𝑠𝑖𝑛𝜃
𝑑𝜃
𝑑𝑦 𝜃
= − tan provided1 + 𝑐𝑜𝑠𝜃 ≠ 0 or 𝜃 ≠ ±𝜋
𝑑𝑥 2

When 𝜃 = 𝜋
The equation of the tangent is 𝑥 = 𝑎𝜋.
When 𝜃 = 0
The equation of the tangent is 𝑦 = 2𝑎.
Similarly the tangents at the other points can be checked and with the above information we can
trace the following:

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Summary
• Steps for curve sketching of a Cartesian curve:
➢ Domain
➢ Intercepts
➢ Symmetry
➢ Asymptotes
➢ Intervals of Increase and Decrease
➢ Local Maximum and Minimum
➢ Concavity/Convexity and Points of Inflection
➢ Graph of the Function
➢ Steps for curve sketching of a polar curve:
➢ Symmetry
➢ Pole
➢ Asymptote
➢ Region
➢ Specific points

Keywords
Curve tracing, Cartesian curve, polar curve, parametric curve

Self Assessment
1. For the curve 𝑥 2 𝑦 2 (𝑥 + 𝑦) = 10 , which of the following is true?
A. The curve has an intercept on x-axis
B. The curve has an intercept on y-axis
C. The curve has an intercept on x-axis and y-axis
D. The curve has no intercept on x-axis and y-axis

2. For the curve 𝑥 2 𝑦 2 (𝑥 + 𝑦) = 10 , which of the following is true?

A. The curve is symmetric about x-axis only
B. The curve is symmetric about y-axis only
C. The curve is symmetric about the line 𝑦 = 𝑥
D. The curve is symmetric about x-axis, y-axis and the line 𝑦 = 𝑥

3. For the curve 𝑥 2 𝑦 2 (𝑥 + 𝑦) = 10

A. there is one asymptote only
B. there are two asymptotes
C. there are three asymptotes
D. there are no asymptotes

4. For the curve 𝑦 = 𝑡𝑎𝑛𝑥 which of the following is false?

A. The origin is a cusp
B. 𝑥 = 2𝑎 is an asymptote
C. Curve exists for all non-negative values of 𝑥

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 Notes
Unit 14: Tracing of Curves
D. The curve is symmetrical about the 𝑥 − 𝑎𝑥𝑖𝑠

5. The folium of Descartes is given by

A. 𝑥 3 + 𝑦 3 = 3𝑎𝑥 2 𝑦 2
B. 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
C. 𝑥 2 + 𝑦 2 = 3𝑎𝑥𝑦
D. 𝑥 4 + 𝑦 4 = 3𝑎𝑥𝑦

6. The curve is symmetric about the polar axis if for every point (r, θ) on the graph, the point
(r, −θ) is also on the graph.
A. True
B. False

7. The curve is symmetric about the polar axis if for some point (r, θ) on the graph, the point
(r, −θ) is also on the graph.
A. True
B. False

8. The curve passes through the pole, if for r = 0, there corresponds a real value of θ.
A. True
B. False

9. The equation 𝑟 = 𝑎(1 + 𝑐𝑜𝑠 𝜃) represents

A. a circle
B. a lemniscate
C. a cardioid
D. a cycloid

10. The equation 𝑝 = 𝑟 𝑐𝑜𝑠(𝜃 − 𝛼) with usual notations, represents a

A. line
B. circle
C. cardioid
D. lemniscate

11. The equations of type x = f(t), y = g(t) are called

A. simultaneous equations
B. ordinary equations
C. parametric equations
D. none of these

12. In the equations of type x = f(t), y = g(t), t is called the parameter.

A. True
B. False

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Notes Calculus

13. The curve having parametric equations 𝑥 = 5 𝑐𝑜𝑠 3 𝜃, 𝑦 = 7 𝑠𝑖𝑛3 𝜃 is

A. a circle
B. an ellipse
C. a cycloid
D. an asteroid

14. The curve 𝑟 = 𝑎 + 𝑏𝑐𝑜𝑠 𝜃 is symmetrical about

A. initial line
B. y-axis
C. line perpendicular to the initial line
𝜋
D. line 𝜃 =
4

15. Number of loops in the curve 𝑟 = 𝑎 𝑐𝑜𝑠 2𝜃 is

A. 2
B. 3
C. 4
D. 6

Answers for Self Assessment

1. D 2. C 3. A 4. D 5. B

6. A 7. B 8. A 9. C 10. A

11. C 12. A 13. D 14. A 15. C

Review Questions
Trace the following curves

1. 𝑦 2 = 8𝑥
2. 𝑥 2 + 𝑦2 = 9
3. 𝑥 2 𝑦 2 (𝑥 + 𝑦) = 10
4. 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦
5. 𝑦 2 𝑥 2 = 𝑥 2 − 𝑎2
6. 𝑟 = 𝑎(1 + 𝑐𝑜𝑠 𝜃)
7. 𝑟 = 𝑎(1 − 𝑐𝑜𝑠 𝜃)
8. 𝑥 = 5 𝑐𝑜𝑠 3 𝜃, 𝑦 = 7 𝑠𝑖𝑛3 𝜃
9. 𝑟 = 𝑎 cos 2𝜃, 𝑎 > 0
10. 𝑟 = 𝑎 + 𝑏𝑐𝑜𝑠 𝜃
11. 𝑥 = 0.5 sec 𝑡, 𝑦 = 1 + cot 𝑡
12. 𝑥 = cos 𝑡, 𝑦 = cot 𝑡
13. 𝑟 log 𝜃 = 𝑎
14. 𝑟 = 𝑎(𝜃 − sin 𝜃)
15. 𝑟 = 𝑎 + 𝑎 sin 𝜃, 𝑎 > 0

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 Notes
Unit 14: Tracing of Curves
Further Reading
W. Thomas Finny (1998). Calculus and Analytic Geometry, 6th Edition, Publishers, Narsa,
India.
R. K. Jain, and Iyengar, SRK. (2010). Advanced Engineering Mathematics, 3 rd Edition
Publishers, Narsa, India.
Widder, D.V. (2002). Advance Calculus 2nd Edition, Publishers, PHI, India.
Piskunov, N. (1996). Differential and Integral Calculus Vol I, & II, Publishers, CBS, India.

https://maths4iitjee.page.tl/Curve-Tracing.htm
https://www.desmos.com/calculator/uu1erqkbey
https://www.kristakingmath.com/blog/sketching-polar-curves
https://brilliant.org/wiki/polar-curves/
https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcen
dentals_(Stewart)/10%3A_Parametric_Equations_And_Polar_Coordinates/10.01%3A_Cu
rves_Defined_by_Parametric_Equations
https://tutorial.math.lamar.edu/classes/calcii/parametriceqn.aspx

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