104

MATHEMATICS

Chapter

CONTINUITY AND
DIFFERENTIABILITY
v The whole of science is nothing more than a refinement
of everyday thinking.” — ALBERT EINSTEIN v
5.1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI. We had learnt to
differentiate certain functions like polynomial functions and
trigonometric functions. In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them. We will also learn differentiation
of inverse trigonometric functions. Further, we introduce a
new class of functions called exponential and logarithmic
functions. These functions lead to powerful techniques of
differentiation. We illustrate certain geometrically obvious
conditions through differential calculus. In the process, we
will learn some fundamental theorems in this area.
Sir Issac Newton
(1642-1727)

5.2 Continuity

We start the section with two informal examples to get a feel of continuity.
Consider
the function
1, if x ≤ 0
f ( x) = 
2, if x > 0

This function is of course defined at every

point of the real line. Graph of this function is
given in the Fig 5.1. One can deduce from the
graph that the value of the function at nearby
points on x-axis remain close to each other
except at x = 0. At the points near and to the
left of 0, i.e., at points like – 0.1, – 0.01, – 0.001,
the value of the function is 1. At the points near
and to the right of 0, i.e., at points like 0.1, 0.01,

Reprint 2024-25

Fig 5.1
CONTINUITY AND DIFFERENTIABILITY

105

0.001, the value of the function is 2. Using the language of left and right hand
limits, we
may say that the left (respectively right) hand limit of f at 0 is 1 (respectively
2). In
particular the left and right hand limits do not coincide. We also observe that the
value
of the function at x = 0 concides with the left hand limit. Note that when we try
to draw
the graph, we cannot draw it in one stroke, i.e., without lifting pen from the
plane of the
paper, we can not draw the graph of this function. In fact, we need to lift the pen
when
we come to 0 from left. This is one instance of function being not continuous at x
= 0.
Now, consider the function defined as
1, if x ≠ 0
f ( x) = 
2, if x = 0

This function is also defined at every point. Left and the right hand limits at x =
0
are both equal to 1. But the value of the
function at x = 0 equals 2 which does not
coincide with the common value of the left
and right hand limits. Again, we note that we
cannot draw the graph of the function without
lifting the pen. This is yet another instance of
a function being not continuous at x = 0.
Naively, we may say that a function is
continuous at a fixed point if we can draw the
graph of the function around that point without
lifting the pen from the plane of the paper.

Fig 5.2

Mathematically, it may be phrased precisely as follows:

Definition 1 Suppose f is a real function on a subset of the real numbers and let c
be
a point in the domain of f. Then f is continuous at c if
lim f ( x ) = f (c )
x→ c

More elaborately, if the left hand limit, right hand limit and the value of the
function
at x = c exist and equal to each other, then f is said to be continuous at x = c.
Recall that
if the right hand and left hand limits at x = c coincide, then we say that the
common
value is the limit of the function at x = c. Hence we may also rephrase the
definition of
continuity as follows: a function is continuous at x = c if the function is defined
at
x = c and if the value of the function at x = c equals the limit of the function at
x = c. If f is not continuous at c, we say f is discontinuous at c and c is called
a point
of discontinuity of f.

Reprint 2024-25
106

MATHEMATICS

Example 1 Check the continuity of the function f given by f (x) = 2x + 3 at x = 1.

Solution First note that the function is defined at the given point x = 1 and its
value is 5.
Then find the limit of the function at x = 1. Clearly
lim f ( x ) = lim (2 x + 3) = 2(1) + 3 = 5
x →1

Thus

x →1

lim f ( x ) = 5 = f (1)
x →1

Hence, f is continuous at x = 1.
Example 2 Examine whether the function f given by f (x) = x2 is continuous at x =
0.
Solution First note that the function is defined at the given point x = 0 and its
value is 0.
Then find the limit of the function at x = 0. Clearly
lim f ( x) = lim x 2 = 02 = 0
x→ 0

Thus

x→ 0

lim f ( x ) = 0 = f (0)
x→ 0

Hence, f is continuous at x = 0.
Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0.
Solution By definition
 − x, if x < 0
f (x) = 
 x, if x ≥ 0
Clearly the function is defined at 0 and f (0) = 0. Left hand limit of f at 0 is

lim f ( x) = lim− (– x) = 0

x→ 0 −

x→ 0

Similarly, the right hand limit of f at 0 is

lim f ( x) = lim+ x = 0

x → 0+

x→ 0

Thus, the left hand limit, right hand limit and the value of the function coincide
at
x = 0. Hence, f is continuous at x = 0.
Example 4 Show that the function f given by
3
 x + 3, if x ≠ 0
f (x) = 
if x = 0
1,

is not continuous at x = 0.

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

107

Solution The function is defined at x = 0 and its value at x = 0 is 1. When x ≠ 0,

the
function is given by a polynomial. Hence,
lim f ( x ) = lim ( x3 + 3) = 03 + 3 = 3
x→ 0

x→0

Since the limit of f at x = 0 does not coincide with f (0), the function is not
continuous
at x = 0. It may be noted that x = 0 is the only point of discontinuity for this
function.
Example 5 Check the points where the constant function f (x) = k is continuous.
Solution The function is defined at all real numbers and by definition, its value
at any
real number equals k. Let c be any real number. Then
lim f ( x ) = lim k = k
x→ c

x→ c

Since f (c) = k = lim

x → c f (x) for any real number c, the function f is continuous at
every real number.
Example 6 Prove that the identity function on real numbers given by f (x) = x is
continuous at every real number.
Solution The function is clearly defined at every point and f (c) = c for every
real
number c. Also,
lim f ( x) = lim x = c
x→c

x→ c

Thus, lim f (x) = c = f (c) and hence the function is continuous at every real
number.
x→ c

Having defined continuity of a function at a given point, now we make a natural

extension of this definition to discuss continuity of a function.
Definition 2 A real function f is said to be continuous if it is continuous at
every point
in the domain of f.
This definition requires a bit of elaboration. Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at
every
point in [a, b] including the end points a and b. Continuity of f at a means

lim f ( x) = f (a)

x→ a+

and continuity of f at b means

lim f ( x) = f(b)
x→b –

Observe that lim− f ( x) and lim+ f ( x) do not make sense. As a consequence

x→ a

x→b

of this definition, if f is defined only at one point, it is continuous there,

i.e., if the
domain of f is a singleton, f is a continuous function.

Reprint 2024-25
108

MATHEMATICS

Example 7 Is the function defined by f (x) = | x |, a continuous function?

Solution We may rewrite f as
 − x, if x < 0
f (x) = 
 x, if x ≥ 0
By Example 3, we know that f is continuous at x = 0.

Let c be a real number such that c < 0. Then f (c) = – c. Also

lim f ( x ) = lim ( − x ) = – c
x→ c

x→c

(Why?)

Since lim f ( x) = f (c) , f is continuous at all negative real numbers.

x→ c

Now, let c be a real number such that c > 0. Then f (c) = c. Also
lim f ( x ) = lim x = c
x→ c

x→ c

(Why?)

Since lim f ( x) = f (c) , f is continuous at all positive real numbers. Hence, f

x→ c

is continuous at all points.

Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 – 1.
Solution Clearly f is defined at every real number c and its value at c is c3 + c2
– 1. We
also know that
lim f ( x ) = lim ( x3 + x 2 − 1) = c3 + c 2 − 1
x→c

x→ c

Thus lim f ( x ) = f (c ) , and hence f is continuous at every real number. This

means
x→ c

f is a continuous function.
Example 9 Discuss the continuity of the function f defined by f (x) =

1
, x ≠ 0.
x

Solution Fix any non zero real number c, we have

1 1
=
x→c x
c

lim f ( x) = lim
x→c

1
, we have lim f ( x ) = f (c ) and hence, f is continuous
x→ c
c
at every point in the domain of f. Thus f is a continuous function.
Also, since for c ≠ 0, f (c) =

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

109

We take this opportunity to explain the concept of infinity. This we do by

analysing
1
the function f (x) = near x = 0. To carry out this analysis we follow the usual
trick of
x
finding the value of the function at real numbers close to 0. Essentially we are
trying to
find the right hand limit of f at 0. We tabulate this in the following (Table 5.1).
Table 5.1

0.3

f (x)

1 3.333...

0.2

0.1 = 10–1

0.01 = 10–2

0.001 = 10–3

10–n

10

100 = 102

1000 = 103

10n

We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher. This may be rephrased as: the value of f (x) may be made larger than any
given
number by choosing a positive real number very close to 0. In symbols, we write

lim f ( x) = + ∞

x → 0+

(to be read as: the right hand limit of f (x) at 0 is plus infinity). We wish to
emphasise
that + ∞ is NOT a real number and hence the right hand limit of f at 0 does not
exist (as
a real number).
Similarly, the left hand limit of f at 0 may be found. The following table is self
explanatory.
Table 5.2

–1

– 0.3

f (x)

– 1 – 3.333...

– 0.2

– 10–1

– 10–2

– 10–3

– 10–n

–5

– 10

– 102

– 103

– 10n

From the Table 5.2, we deduce that the

value of f (x) may be made smaller than any
given number by choosing a negative real
number very close to 0. In symbols,
we write

lim f ( x) = − ∞

x → 0−

(to be read as: the left hand limit of f (x) at 0 is

minus infinity). Again, we wish to emphasise
that – ∞ is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number). The graph of the reciprocal function
given in Fig 5.3 is a geometric representation
of the above mentioned facts.

Reprint 2024-25

Fig 5.3
110

MATHEMATICS

Example 10 Discuss the continuity of the function f defined by

 x + 2, if x ≤ 1
f (x) = 
 x − 2, if x > 1

Solution The function f is defined at all points of the real line.

Case 1 If c < 1, then f (c) = c + 2. Therefore, lim f ( x ) = lim( x + 2) = c + 2
x→ c

x→c

Thus, f is continuous at all real numbers less than 1.

Case 2 If c > 1, then f (c) = c – 2. Therefore,
lim f ( x) = lim (x – 2) = c – 2 = f (c)
x →c

x →c

Thus, f is continuous at all points x > 1.

Case 3 If c = 1, then the left hand limit of f at
x = 1 is
lim f ( x) = lim– ( x + 2) = 1 + 2 = 3

x →1–

x →1

The right hand limit of f at x = 1 is

lim f ( x) = lim+ ( x − 2) = 1 − 2 = −1

x→1+

x →1

Since the left and right hand limits of f at x = 1

Fig 5.4
do not coincide, f is not continuous at x = 1. Hence
x = 1 is the only point of discontinuity of f. The graph of the function is given
in Fig 5.4.
Example 11 Find all the points of discontinuity of the function f defined by
 x + 2, if x < 1

f (x) =  0, if x = 1
 x − 2, if x > 1

Solution As in the previous example we find that f
is continuous at all real numbers x ≠ 1. The left
hand limit of f at x = 1 is
lim− f ( x) = lim– ( x + 2) = 1 + 2 = 3
x →1

x →1

The right hand limit of f at x = 1 is

lim+ f ( x) = lim+ ( x − 2) = 1 − 2 = −1
x→1

x →1

Since, the left and right hand limits of f at x = 1

do not coincide, f is not continuous at x = 1. Hence
x = 1 is the only point of discontinuity of f. The
graph of the function is given in the Fig 5.5.

Reprint 2024-25

Fig 5.5
CONTINUITY AND DIFFERENTIABILITY

111

Example 12 Discuss the continuity of the function defined by

 x + 2, if x < 0
f (x) = 
 − x + 2, if x > 0

Solution Observe that the function is defined at all real numbers except at 0.
Domain
of definition of this function is
D1 ∪ D2 where D1 = {x ∈ R : x < 0} and
D2 = {x ∈ R : x > 0}
Case 1 If c ∈ D 1, then lim f ( x) = lim (x + 2)
x →c

x →c

= c + 2 = f (c) and hence f is continuous in D1.

Case 2 If c ∈ D2, then lim f ( x) = lim (– x + 2)
x →c

x →c

= – c + 2 = f (c) and hence f is continuous in D2.

Since f is continuous at all points in the domain of f,
we deduce that f is continuous. Graph of this
function is given in the Fig 5.6. Note that to graph
Fig 5.6
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is
not
defined.
Example 13 Discuss the continuity of the function f given by
 x, if x ≥ 0
f (x) =  2
 x , if x < 0

Solution Clearly the function is defined at

every real number. Graph of the function is
given in Fig 5.7. By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line.
Let

D1 = {x ∈ R : x < 0}, D2 = {0} and

D3 = {x ∈ R : x > 0}

Fig 5.7

Case 1 At any point in D1, we have f (x) = x2 and it is easy to see that it is
continuous
there (see Example 2).
Case 2 At any point in D3, we have f (x) = x and it is easy to see that it is
continuous
there (see Example 6).

Reprint 2024-25
112

MATHEMATICS

Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f

(0) = 0.
The left hand limit of f at 0 is

lim f ( x) = lim− x 2 = 02 = 0

x →0 –

x →0

The right hand limit of f at 0 is

lim f ( x) = lim+ x = 0

x → 0+

x →0

Thus lim f ( x) = 0 = f (0) and hence f is continuous at 0. This means that f is

x →0

continuous at every point in its domain and hence, f is a continuous function.

Example 14 Show that every polynomial function is continuous.
Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x + ... + an xn for some natural number n, an ≠ 0 and ai ∈ R.
Clearly this
function is defined for every real number. For a fixed real number c, we have
lim p ( x) = p (c)
x →c

By definition, p is continuous at c. Since c is any real number, p is continuous at

every real number and hence p is a continuous function.
Example 15 Find all the points of discontinuity of the greatest integer function
defined
by f (x) = [x], where [x] denotes the greatest integer less than or equal to x.
Solution First observe that f is defined for all real numbers. Graph of the
function is
given in Fig 5.8. From the graph it looks like that f is discontinuous at every
integral
point. Below we explore, if this is true.

Fig 5.8

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

113

Case 1 Let c be a real number which is not equal to any integer. It is evident from
the
graph that for all real numbers close to c the value of the function is equal to
[c]; i.e.,
lim f ( x) = lim [ x] = [c] . Also f (c) = [c] and hence the function is continuous
at all real
x →c

x →c

numbers not equal to integers.

Case 2 Let c be an integer. Then we can find a sufficiently small real number
r > 0 such that [c – r] = c – 1 whereas [c + r] = c.
This, in terms of limits mean that
lim f (x) = c – 1, lim+ f (x) = c

x →c −

x →c

Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point.
5.2.1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt
some
algebra of limits. Analogously, now we will study some algebra of continuous
functions.
Since continuity of a function at a point is entirely dictated by the limit of the
function at
that point, it is reasonable to expect results analogous to the case of limits.
Theorem 1 Suppose f and g be two real functions continuous at a real number c.
Then
(1) f + g is continuous at x = c.
(2) f – g is continuous at x = c.
(3) f . g is continuous at x = c.
 f 
(4)   is continuous at x = c, (provided g (c) ≠ 0).
g

Proof We are investigating continuity of (f + g) at x = c. Clearly it is defined at

x = c. We have
lim( f + g ) ( x ) = lim [ f ( x) + g ( x)]
x →c

x →c

(by definition of f + g)

= lim f ( x) + lim g ( x)

(by the theorem on limits)

= f (c) + g(c)
= (f + g) (c)
(as f and g are continuous)
(by definition of f + g)

x →c

x →c

Hence, f + g is continuous at x = c.
Proofs for the remaining parts are similar and left as an exercise to the reader.

Reprint 2024-25
114

MATHEMATICS

Remarks
(i) As a special case of (3) above, if f is a constant function, i.e., f (x) = λ
for some
real number λ, then the function (λ . g) defined by (λ . g) (x) = λ . g(x) is also
continuous. In particular if λ = – 1, the continuity of f implies continuity of –
f.
(ii) As a special case of (4) above, if f is the constant function f (x) = λ, then
the

λ
λ
λ
defined by ( x) =
is also continuous wherever g (x) ≠ 0. In
g
g
g ( x)
1
particular, the continuity of g implies continuity of .
g
The above theorem can be exploited to generate many continuous functions. They
also aid in deciding if certain functions are continuous or not. The following
examples
illustrate this:
function

Example 16 Prove that every rational function is continuous.

Solution Recall that every rational function f is given by

p ( x)
, q ( x) ≠ 0
q( x)
where p and q are polynomial functions. The domain of f is all real numbers except
points at which q is zero. Since polynomial functions are continuous (Example 14),
f is
continuous by (4) of Theorem 1.
f ( x) =

Example 17 Discuss the continuity of sine function.

Solution To see this we use the following facts
lim sin x = 0
x →0

We have not proved it, but is intuitively clear from the graph of sin x near 0.
Now, observe that f (x) = sin x is defined for every real number. Let c be a real
number. Put x = c + h. If x → c we know that h → 0. Therefore
lim f ( x) = lim sin x
x →c

x→c

sin(c + h)
= lim
h →0
[sin c cos h + cos c sin h]
= lim
h →0
[sin c cos h] + lim [cos c sin h]
= lim
h →0
h →0

= sin c + 0 = sin c = f (c)

Thus lim f (x) = f (c) and hence f is a continuous function.
x →c

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

115

Remark A similar proof may be given for the continuity of cosine function.
Example 18 Prove that the function defined by f (x) = tan x is a continuous
function.
Solution The function f (x) = tan x =

sin x
. This is defined for all real numbers such
cos x

π
. We have just proved that both sine and cosine
2
functions are continuous. Thus tan x being a quotient of two continuous functions
is
continuous wherever it is defined.
An interesting fact is the behaviour of continuous functions with respect to
composition of functions. Recall that if f and g are two real functions, then
(f o g) (x) = f (g (x))
is defined whenever the range of g is a subset of domain of f. The following
theorem
(stated without proof) captures the continuity of composite functions.
Theorem 2 Suppose f and g are real valued functions such that (f o g) is defined at
c.
If g is continuous at c and if f is continuous at g (c), then (f o g) is continuous
at c.
The following examples illustrate this theorem.
that cos x ≠ 0, i.e., x ≠ (2n +1)

Example 19 Show that the function defined by f (x) = sin (x2) is a continuous
function.
Solution Observe that the function is defined for every real number. The function
f may be thought of as a composition g o h of the two functions g and h, where
g (x) = sin x and h (x) = x2. Since both g and h are continuous functions, by
Theorem 2,
it can be deduced that f is a continuous function.
Example 20 Show that the function f defined by
f (x) = |1 – x + | x | |,
where x is any real number, is a continuous function.
Solution Define g by g (x) = 1 – x + | x | and h by h (x) = | x | for all real x.
Then
(h o g) (x) = h (g (x))
= h (1– x + | x |)
= | 1– x + | x | | = f (x)
In Example 7, we have seen that h is a continuous function. Hence g being a sum
of a polynomial function and the modulus function is continuous. But then f being a
composite of two continuous functions is continuous.

Reprint 2024-25
116

MATHEMATICS

EXERCISE 5.1
1. Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at
x = 5.
2. Examine the continuity of the function f (x) = 2x2 – 1 at x = 3.
3. Examine the following functions for continuity.
(a) f (x) = x – 5

(b) f (x) =

1
,x≠5
x −5

x 2 − 25
, x ≠ –5
(d) f (x) = | x – 5 |
x+5
4. Prove that the function f (x) = xn is continuous at x = n, where n is a positive
integer.
5. Is the function f defined by

(c) f (x) =

 x, if x ≤ 1
f ( x) = 
5, if x > 1
continuous at x = 0? At x = 1? At x = 2?
Find all points of discontinuity of f, where f is defined by

7.

| x | +3, if x ≤ − 3

f ( x) =  −2 x, if − 3 < x < 3
6 x + 2, if x ≥ 3


8.

| x |
 , if x ≠ 0
f ( x) =  x
 0, if x = 0

9.

 x
 , if x < 0
f ( x) =  | x |
 −1, if x ≥ 0


10.

 x + 1, if x ≥ 1
f ( x) =  2
 x + 1, if x < 1

11.

12.

10
 x − 1, if x ≤ 1
f ( x) =  2
if x > 1
 x ,

6.

2 x + 3, if x ≤ 2
f ( x) = 
2 x − 3, if x > 2

3
 x − 3, if x ≤ 2
f ( x) =  2
 x + 1, if x > 2

13. Is the function defined by

 x + 5, if x ≤ 1
f ( x) = 
 x − 5, if x > 1

a continuous function?

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

117

Discuss the continuity of the function f, where f is defined by

14.

3, if 0 ≤ x ≤ 1

f ( x) =  4, if 1 < x < 3
5, if 3 ≤ x ≤ 10


15.

2 x, if x < 0

f ( x) = 0, if 0 ≤ x ≤ 1
4 x, if x > 1


−2, if x ≤ − 1

16. f ( x) =  2 x, if − 1 < x ≤ 1
2, if x > 1

17. Find the relationship between a and b so that the function f defined by
ax + 1, if x ≤ 3
f ( x) = 
bx + 3, if x > 3

is continuous at x = 3.
18. For what value of λ is the function defined by

19.
20.
21.

22.
23.

λ ( x 2 − 2 x), if x ≤ 0
f ( x) = 
if x > 0
4 x + 1,
continuous at x = 0? What about continuity at x = 1?
Show that the function defined by g (x) = x – [x] is discontinuous at all integral
points. Here [x] denotes the greatest integer less than or equal to x.
Is the function defined by f (x) = x2 – sin x + 5 continuous at x = π?
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
(b) f (x) = sin x – cos x
(c) f (x) = sin x . cos x
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Find all points of discontinuity of f, where

 sin x
, if x < 0

f ( x) =  x
 x + 1, if x ≥ 0
24. Determine if f defined by
1
 2
 x sin , if x ≠ 0
f ( x) = 
x
0,
if x = 0
is a continuous function?

Reprint 2024-25
118

MATHEMATICS

25. Examine the continuity of f, where f is defined by

sin x − cos x, if x ≠ 0
f ( x) = 
if x = 0
 −1,
Find the values of k so that the function f is continuous at the indicated point in
Exercises
26 to 29.

26.

π
 k cos x
 π − 2 x , if x ≠ 2
f ( x) = 
π
3,
if x =

2

at x =

27.

2
kx , if x ≤ 2
f ( x) = 
if x > 2
3,

at x = 2

28.

 kx + 1, if x ≤ π
f ( x) = 
cos x, if x > π

at x = π

π
2

 kx + 1, if x ≤ 5
f ( x) = 
at x = 5
3x − 5, if x > 5
30. Find the values of a and b such that the function defined by

29.

31.
32.
33.
34.

if x ≤ 2
5,

f ( x) = ax + b, if 2 < x < 10
21,
if x ≥ 10

is a continuous function.
Show that the function defined by f (x) = cos (x2) is a continuous function.
Show that the function defined by f (x) = | cos x | is a continuous function.
Examine that sin | x | is a continuous function.
Find all the points of discontinuity of f defined by f (x) = | x | – | x + 1 |.

5.3. Differentiability
Recall the following facts from previous class. We had defined the derivative of a
real
function as follows:
Suppose f is a real function and c is a point in its domain. The derivative of f at
c is
defined by

lim

h →0

f (c + h ) − f (c )
h

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

provided this limit exists. Derivative of f at c is denoted by f ′(c) or

119

d
( f ( x)) | c . The
dx

function defined by

f ′( x) = lim

h →0

f ( x + h) − f ( x )
h

wherever the limit exists is defined to be the derivative of f. The derivative of f

is

d
dy
( f ( x)) or if y = f (x) by
or y′. The process of finding
dx
dx
derivative of a function is called differentiation. We also use the phrase
differentiate
f (x) with respect to x to mean find f ′(x).
The following rules were established as a part of algebra of derivatives:
(1) (u ± v)′ = u′ ± v′
(2) (uv)′ = u′v + uv′ (Leibnitz or product rule)
denoted by f ′ (x) or

′
(3)  u  = u′v − uv′ , wherever v ≠ 0 (Quotient rule).
v
v2
The following table gives a list of derivatives of certain standard functions:
Table 5.3

f (x)

xn

sin x

cos x

tan x

f ′(x)

nx n – 1

cos x
– sin x

sec2 x

Whenever we defined derivative, we had put a caution provided the limit exists.
Now the natural question is; what if it doesn’t? The question is quite pertinent
and so is

f (c + h ) − f (c )
does not exist, we say that f is not differentiable at c.
h
In other words, we say that a function f is differentiable at a point c in its
domain if both
its answer. If lim

h →0

f (c + h ) − f ( c )
f (c + h ) − f (c )
and lim+
are finite and equal. A function is said
h →0
h →0
h
h
to be differentiable in an interval [a, b] if it is differentiable at every point
of [a, b]. As
in case of continuity, at the end points a and b, we take the right hand limit and
left hand
limit, which are nothing but left hand derivative and right hand derivative of the
function
at a and b respectively. Similarly, a function is said to be differentiable in an
interval
(a, b) if it is differentiable at every point of (a, b).
lim–

Reprint 2024-25
120

MATHEMATICS

Theorem 3 If a function f is differentiable at a point c, then it is also

continuous at that
point.
Proof Since f is differentiable at c, we have

f ( x ) − f (c )
= f ′(c)
x →c
x−c

lim
But for x ≠ c, we have

f (x) – f (c) =
Therefore

or

f ( x ) − f (c )
. ( x − c)
x−c

 f ( x ) − f (c )

. ( x − c) 
lim [ f ( x) − f (c)] = lim 
x →c 
x →c
x−c


f ( x ) − f (c ) 
lim [ f ( x)] − lim [ f (c)] = lim 
. lim [( x − c)]

x →c
x →c
x →c 
x−c
 x →c
= f ′(c) . 0 = 0

lim f ( x) = f (c)

or

x →c

Hence f is continuous at x = c.
Corollary 1 Every differentiable function is continuous.
We remark that the converse of the above statement is not true. Indeed we have
seen that the function defined by f (x) = | x | is a continuous function. Consider
the left
hand limit
lim–

h →0

f (0 + h) − f (0) −h
=
= −1
h
h

The right hand limit

lim+

h →0

f (0 + h) − f (0) h
= =1
h
h

f (0 + h) − f (0)
h
does not exist and hence f is not differentiable at 0. Thus f is not a
differentiable
function.
Since the above left and right hand limits at 0 are not equal, lim

h →0

5.3.1 Derivatives of composite functions

To study derivative of composite functions, we start with an illustrative example.
Say,
we want to find the derivative of f, where
f (x) = (2x + 1)3

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

121

One way is to expand (2x + 1)3 using binomial theorem and find the derivative as
a polynomial function as illustrated below.

d
d
(2 x + 1)3 
f ( x) =
dx
dx
d
(8 x 3 + 12 x 2 + 6 x + 1)
dx
= 24x2 + 24x + 6
= 6 (2x + 1)2
Now, observe that
f (x) = (h o g) (x)
where g(x) = 2x + 1 and h(x) = x3. Put t = g(x) = 2x + 1. Then f(x) = h(t) = t3.
Thus
=

df
dh dt
⋅
= 6 (2x + 1)2 = 3(2x + 1)2 . 2 = 3t2 . 2 =
dx
dt dx
The advantage with such observation is that it simplifies the calculation in
finding
the derivative of, say, (2x + 1)100. We may formalise this observation in the
following
theorem called the chain rule.
Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two
functions u and v ; i.e., f = v o u. Suppose t = u (x) and if both

dt
dv
and
exist, we have
dx
dt

df dv dt
= ⋅
dx dt dx
We skip the proof of this theorem. Chain rule may be extended as follows. Suppose
f is a real valued function which is a composite of three functions u, v and w ;
i.e.,
f = (w o u) o v. If t = v (x) and s = u (t), then
df d ( w o u ) dt dw ds dt
=
⋅
=
⋅ ⋅
dx
dt
dx ds dt dx
provided all the derivatives in the statement exist. Reader is invited to formulate
chain
rule for composite of more functions.
Example 21 Find the derivative of the function given by f (x) = sin (x2).
Solution Observe that the given function is a composite of two functions. Indeed,
if
t = u(x) = x2 and v(t) = sin t, then
f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2

Reprint 2024-25
122

MATHEMATICS

Put t = u(x) = x2. Observe that

dv
dt
= cos t and
= 2 x exist. Hence, by chain rule
dt
dx

df
dv dt
⋅ = cos t ⋅ 2 x
=
dx
dt dx
It is normal practice to express the final result only in terms of x. Thus
df
= cos t ⋅ 2 x = 2 x cos x 2
dx

EXERCISE 5.2
Differentiate the functions with respect to x in Exercises 1 to 8.
1. sin (x2 + 5)
2. cos (sin x)
3. sin (ax + b)
4. sec (tan ( x ))

sin (ax + b)
5. cos (cx + d )

6. cos x3 . sin2 (x5)

7. 2 cot ( x 2 )
8. cos ( x )
9. Prove that the function f given by
f (x) = | x – 1 |, x ∈ R
is not differentiable at x = 1.
10. Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2.
5.3.2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f
(x).
But it is not necessary that functions are always expressed in this form. For
example,
consider one of the following relationships between x and y:
x–y–π=0
x + sin xy – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π. In
the second case, it does not seem that there is an easy way to solve for y.
Nevertheless,
there is no doubt about the dependence of y on x in either of the cases. When a
relationship between x and y is expressed in a way that it is easy to solve for y
and
write y = f (x), we say that y is given as an explicit function of x. In the latter
case it

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

123

is implicit that y is a function of x and we say that the relationship of the

second type,
above, gives function implicitly. In this subsection, we learn to differentiate
implicit
functions.
Example 22 Find

dy
if x – y = π.
dx

Solution One way is to solve for y and rewrite the above as

y=x–π

dy
=1
dx
Alternatively, directly differentiating the relationship w.r.t., x, we have

But then

d
dπ
( x − y) =
dx
dx
dπ
means to differentiate the constant function taking value π
dx
everywhere w.r.t., x. Thus
Recall that

d
d
( x) − ( y ) = 0
dx
dx
which implies that

dy
dx
=1
=
dx
dx
Example 23 Find

dy
, if y + sin y = cos x.
dx

Solution We differentiate the relationship directly with respect to x, i.e.,

dy d
d
+ (sin y ) =
(cos x)
dx dx
dx
which implies using chain rule

dy
dy
+ cos y ⋅
= – sin x
dx
dx
This gives

dy
sin x
= −
dx
1 + cos y

where

y ≠ (2n + 1) π

Reprint 2024-25
124

MATHEMATICS

5.3.3 Derivatives of inverse trigonometric functions

We remark that inverse trigonometric functions are continuous functions, but we
will
not prove this. Now we use chain rule to find derivatives of these functions.
Example 24 Find the derivative of f given by f (x) = sin–1 x assuming it exists.
Solution Let y = sin–1 x. Then, x = sin y.
Differentiating both sides w.r.t. x, we get
1 = cos y

dy
dx

dy
1
1
=
=
dx
cos y cos (sin −1 x)

which implies that

π π
Observe that this is defined only for cos y ≠ 0, i.e., sin–1 x ≠ − , , i.e., x ≠ –
1, 1,
2 2
i.e., x ∈ (– 1, 1).
To make this result a bit more attractive, we carry out the following manipulation.
Recall that for x ∈ (– 1, 1), sin (sin–1 x) = x and hence
cos2 y = 1 – (sin y)2 = 1 – (sin (sin–1 x))2 = 1 – x2
 π π
Also, since y ∈  − ,  , cos y is positive and hence cos y = 1 − x 2
 2 2
Thus, for x ∈ (– 1, 1),

dy
1
1
=
=
dx cos y
1 − x2

f (x)

sin–1 x

f 1(x)

Domain off

1 − x2
(-1, 1)

cos-1 x
−1

1 − x2

(-1, 1)

Reprint 2024-25

tan-1x

1
1 + x2
R
CONTINUITY AND DIFFERENTIABILITY

125

EXERCISE 5.3
dy
in the following:
dx
1. 2x + 3y = sin x
4. xy + y2 = tan x + y

2. 2x + 3y = sin y
5. x2 + xy + y2 = 100

3. ax + by2 = cos y
6. x3 + x2y + xy2 + y3 = 81

7. sin2 y + cos xy = κ

8. sin2 x + cos2 y = 1

 2x 
9. y = sin–1 

 1 + x2 

Find

1
1
 3 x − x3 
<x<
, −
10. y = tan–1 
2 
3
3
 1 − 3x 

11.

 1 − x2 
y = cos −1 
, 0 < x < 1
 1 + x2 

12.

 1 − x2 
y = sin −1 
, 0 < x < 1
 1 + x2 

13.

 2x  ,
y = cos −1 
 −1 < x < 1
 1 + x2 
14.

y = sin −1 2 x 1 − x 2 , −

15.

1
 1 ,
y = sec −1  2
 0< x<
 2x −1 
2

1
1
<x<
2
2

5.4 Exponential and Logarithmic Functions

Till now we have learnt some aspects of different classes of functions like
polynomial
functions, rational functions and trigonometric functions. In this section, we
shall
learn about a new class of (related) functions called exponential functions and
logarithmic
functions. It needs to be emphasized that many statements made in this section are
motivational and precise proofs of these are well beyond the scope of this text.
The Fig 5.9 gives a sketch of y = f1(x) = x, y = f2(x) = x2, y = f3(x) = x3 and y =
f4(x)
= x . Observe that the curves get steeper as the power of x increases. Steeper the
curve, faster is the rate of growth. What this means is that for a fixed increment
in the
value of x (> 1), the increment in the value of y = fn (x) increases as n increases
for n
= 1, 2, 3, 4. It is conceivable that such a statement is true for all positive
values of n,
4

Reprint 2024-25
126

MATHEMATICS

where fn (x) = xn. Essentially, this means

that the graph of y = fn (x) leans more
towards the y-axis as n increases. For
example, consider f10(x) = x10 and f15(x)
= x 15 . If x increases from 1 to 2, f 10
increases from 1 to 2 10 whereas f 15
increases from 1 to 215. Thus, for the same
increment in x, f15 grow faster than f10.
Upshot of the above discussion is that
the growth of polynomial functions is
dependent on the degree of the polynomial
function – higher the degree, greater is
Fig 5.9
the growth. The next natural question is:
Is there a function which grows faster than any polynomial function. The answer is
in
affirmative and an example of such a function is
y = f (x) = 10x.
Our claim is that this function f grows faster than fn (x) = xn for any positive
integer n.
For example, we can prove that 10x grows faster than f100 (x) = x100. For large
values
3

of x like x = 103, note that f100 (x) = (103)100 = 10300 whereas f (103) = 1010 =
101000.
Clearly f (x) is much greater than f100 (x). It is not difficult to prove that for
all
x > 103, f (x) > f100 (x). But we will not attempt to give a proof of this here.
Similarly, by
choosing large values of x, one can verify that f (x) grows faster than fn (x) for
any
positive integer n.
Definition 3 The exponential function with positive base b > 1 is the function
y = f (x) = bx
The graph of y = 10x is given in the Fig 5.9.
It is advised that the reader plots this graph for particular values of b like 2, 3
and 4.
Following are some of the salient features of the exponential functions:
(1) Domain of the exponential function is R, the set of all real numbers.
(2) Range of the exponential function is the set of all positive real numbers.
(3) The point (0, 1) is always on the graph of the exponential function (this is a
restatement of the fact that b0 = 1 for any real b > 1).
(4) Exponential function is ever increasing; i.e., as we move from left to right,
the
graph rises above.

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

127

(5) For very large negative values of x, the exponential function is very close to
0. In
other words, in the second quadrant, the graph approaches x-axis (but never
meets it).
Exponential function with base 10 is called the common exponential function. In
the Appendix A.1.4 of Class XI, it was observed that the sum of the series

1 1
+ + ...
1! 2!
is a number between 2 and 3 and is denoted by e. Using this e as the base we obtain
an
extremely important exponential function y = ex.
This is called natural exponential function.
It would be interesting to know if the inverse of the exponential function exists
and
has nice interpretation. This search motivates the following definition.

1+

Definition 4 Let b > 1 be a real number. Then we say logarithm of a to base b is x

if
bx = a.
Logarithm of a to base b is denoted by logb a. Thus logb a = x if bx = a. Let us
work with a few explicit examples to get a feel for this. We know 23 = 8. In terms
of
logarithms, we may rewrite this as log2 8 = 3. Similarly, 104 = 10000 is equivalent
to
saying log10 10000 = 4. Also, 625 = 54 = 252 is equivalent to saying log5 625 = 4
or
log25 625 = 2.
On a slightly more mature note, fixing a base b > 1, we may look at logarithm as
a function from positive real numbers to all real numbers. This function, called
the
logarithmic function, is defined by
logb : R+ → R
x → logb x = y if by = x
As before if the base b = 10, we say it
is common logarithms and if b = e, then
we say it is natural logarithms. Often
natural logarithm is denoted by ln. In this
chapter, log x denotes the logarithm
function to base e, i.e., ln x will be written
as simply log x. The Fig 5.10 gives the plots
of logarithm function to base 2, e and 10.
Some of the important observations
about the logarithm function to any base
b > 1 are listed below:

Reprint 2024-25

Fig 5.10
128

MATHEMATICS

(1) We cannot make a meaningful definition of logarithm of non-positive numbers

and hence the domain of log function is R+.
(2) The range of log function is the set of all real numbers.
(3) The point (1, 0) is always on the graph of the log function.
(4) The log function is ever increasing,
i.e., as we move from left to right
the graph rises above.
(5) For x very near to zero, the value
of log x can be made lesser than
any given real number. In other
words in the fourth quadrant the
graph approaches y-axis (but
never meets it).
(6) Fig 5.11 gives the plot of y = ex and
y = ln x. It is of interest to observe
that the two curves are the mirror
images of each other reflected in the line y = x.

Fig 5.11

Two properties of ‘log’ functions are proved below:

(1) There is a standard change of base rule to obtain loga p in terms of logb p.
Let
loga p = α, logb p = β and logb a = γ. This means aα = p, bβ = p and bγ = a.
Substituting the third equation in the first one, we have
(bγ)α = bγα = p
Using this in the second equation, we get
bβ = p = bγα
which implies

β = αγ or α =

loga p =

β
. But then
γ

log b p
log b a

(2) Another interesting property of the log function is its effect on products. Let
logb pq = α. Then bα = pq. If logb p = β and logb q = γ, then bβ = p and bγ = q.
But then bα = pq = bβbγ = bβ + γ
which implies α = β + γ, i.e.,
logb pq = logb p + logb q

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

129

A particularly interesting and important consequence of this is when p = q. In

this case the above may be rewritten as
logb p2 = logb p + logb p = 2 log p
An easy generalisation of this (left as an exercise!) is
logb pn = n log p
for any positive integer n. In fact this is true for any real number n, but we will
not attempt to prove this. On the similar lines the reader is invited to verify

log b

x
= logb x – logb y
y

Example 25 Is it true that x = elog x for all real x?

Solution First, observe that the domain of log function is set of all positive real
numbers.
So the above equation is not true for non-positive real numbers. Now, let y = elog
x. If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x . log e =
log x. Thus
y = x. Hence x = elog x is true only for positive values of x.
One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation. This is
captured
in the following theorem whose proof we skip.
Theorem 5*
(1) The derivative of ex w.r.t., x is ex; i.e.,
(2) The derivative of log x w.r.t., x is

d x
(e ) = ex.
dx

d
1
1
; i.e.,
(log x) = .
dx
x
x

Example 26 Differentiate the following w.r.t. x:

(i) e –x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)

(iv)

Solution
(i) Let y = e – x. Using chain rule, we have

dy
−x d
= e ⋅
(– x) = – e– x
dx
dx
(ii) Let y = sin (log x). Using chain rule, we have
dy
d
cos (log x)
= cos (log x) ⋅ (log x) =
dx
dx
x
* Please see supplementary material on Page 222.

Reprint 2024-25

ecos x
130

MATHEMATICS

(iii) Let y = cos–1 (ex). Using chain rule, we have

−1

dy
=
dx

1 − (e x ) 2

d x
−e x
(e ) =
dx
1 − e2 x

(iv) Let y = ecos x. Using chain rule, we have

dy
= ecos x ⋅ (− sin x) = − (sin x) ecos x
dx

EXERCISE 5.4
Differentiate the following w.r.t. x:
1.

ex
sin x

2. esin

4. sin (tan–1 e–x)

7.

e x, x>0

−1

3. e x

5. log (cos ex)

x
x
x
6. e + e + ... + e

8. log (log x), x > 1

cos x
9. log x , x > 0

10. cos (log x + ex), x > 0

5.5. Logarithmic Differentiation

In this section, we will learn to differentiate certain special class of functions
given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
Using chain rule we may differentiate this to get

1 dy
1 .
u′(x) + v′(x) . log [u(x)]
⋅
= v( x) ⋅
y dx
u ( x)
which implies that
dy
 v( x)

= y
⋅ u′( x) + v′( x) ⋅ log [ u ( x)]
dx
u
(
x
)



The main point to be noted in this method is that f (x) and u(x) must always be
positive as otherwise their logarithms are not defined. This process of
differentiation is
known as logarithms differentiation and is illustrated by the following examples:

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

Example 27 Differentiate

Solution Let y =

( x − 3) ( x 2 + 4)
w.r.t. x.
3x 2 + 4 x + 5

( x − 3) ( x 2 + 4)
(3 x 2 + 4 x + 5)

Taking logarithm on both sides, we have

1
[log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
2
Now, differentiating both sides w.r.t. x, we get
log y =

1 1
2x
6x + 4 
1 dy
+ 2
− 2
⋅
= 
2  ( x − 3) x + 4 3x + 4 x + 5 
y dx

or

2x
6x + 4 
y 1
dy
+ 2
− 2
= 
2  ( x − 3) x + 4 3 x + 4 x + 5 
dx
1
=
2

( x − 3) ( x 2 + 4)  1
2x
6x + 4 
+ 2
− 2
2

3x + 4 x + 5  ( x − 3) x + 4 3 x + 4 x + 5 

Example 28 Differentiate ax w.r.t. x, where a is a positive constant.

Solution Let y = ax. Then
log y = x log a
Differentiating both sides w.r.t. x, we have

1 dy
y dx = log a
or

dy
= y log a
dx

Thus

d x
(a ) = ax log a
dx

Alternatively

d x log a
d
d x
(a ) =
(e
) = e x log a ( x log a )
dx
dx
dx
= ex log a . log a = ax log a.

Reprint 2024-25

131
132

MATHEMATICS

Example 29 Differentiate xsin x, x > 0 w.r.t. x.

Solution Let y = xsin x. Taking logarithm on both sides, we have
log y = sin x log x
Therefore

1 dy
d
d
.
sin x (log x) + log x (sin x)
y dx =
dx
dx

or

1 dy
1
(sin x) + log x cos x
y dx =
x

or

dy
 sin x

+ cos x log x 
= y
 x

dx
= x

sin x  sin x


 x + cos x log x 

= xsin x −1 ⋅ sin x + xsin x ⋅ cos x log x

Example 30 Find

dy
, if yx + xy + xx = ab.
dx

Solution Given that yx + xy + xx = ab.

Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore

du dv dw
+
+
=0
dx dx dx
... (1)

Now, u = yx. Taking logarithm on both sides, we have

log u = x log y
Differentiating both sides w.r.t. x, we have

d
d
1 du
⋅
= x (log y ) + log y ( x )
u dx
dx
dx
1 dy
= x ⋅ + log y ⋅ 1
y dx
So

 x dy

 x dy

du
+ log y  = y x 
+ log y 
= u
dx
 y dx

 y dx


Also v = xy

Reprint 2024-25

... (2)
CONTINUITY AND DIFFERENTIABILITY

133

Taking logarithm on both sides, we have

log v = y log x
Differentiating both sides w.r.t. x, we have

1 dv
d
dy
⋅
= y (log x) + log x
v dx
dx
dx
dy
1
= y ⋅ + log x ⋅
x
dx
dy 
dv
y
= v  + log x 
x
dx 

dx

So

dy 
y y
= x  + log x 
dx 
x
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x.
Differentiating both sides w.r.t. x, we have

... (3)

d
d
1 dw
⋅
= x (log x) + log x ⋅ ( x)
w dx
dx
dx
1
= x ⋅ + log x ⋅ 1
x

dw
= w (1 + log x)
dx

i.e.

= xx (1 + log x)
From (1), (2), (3), (4), we have
dy 
 x dy

y
yx 
+ log y  + x y  + log x  + xx (1 + log x) = 0
dx 
x
 y dx


or
Therefore

(x . yx – 1 + xy . log x)

dy
= – xx (1 + log x) – y . xy–1 – yx log y
dx
− [ y x log y + y . x y −1 + x x (1 + log x)]
dy
=
x . y x −1 + x y log x
dx

Reprint 2024-25

... (4)
134

MATHEMATICS

EXERCISE 5.5
Differentiate the functions given in Exercises 1 to 11 w.r.t. x.

( x − 1) ( x − 2)
( x − 3) ( x − 4) ( x − 5)

1. cos x . cos 2x . cos 3x

2.

3. (log x)cos x

4. xx – 2sin x


1

5. (x + 3)2 . (x + 4)3 . (x + 5)4

1+ 
1

6.  x +  + x x 
x


7. (log x)x + xlog x

8. (sin x)x + sin–1

9. xsin x + (sin x)cos x

10.

x x cos x +

x2 + 1
x2 − 1

11. (x cos x)x + ( x sin x) x

dy
of the functions given in Exercises 12 to 15.
dx
12. xy + yx = 1
13. yx = xy
14. (cos x)y = (cos y)x
15. xy = e(x – y)
16. Find the derivative of the function given by f (x) = (1 + x) (1 + x2) (1 + x4)
(1 + x8)
and hence find f ′(1).
17. Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
18. If u, v and w are functions of x, then show that

Find

d
du
dv
dw
(u. v. w) =
v. w + u .
.w+u.v
dx
dx
dx
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation.

5.6 Derivatives of Functions in Parametric Forms

Sometimes the relation between two variables is neither explicit nor implicit, but
some
link of a third variable with each of the two variables, separately, establishes a
relation
between the first two variables. In such a situation, we say that the relation
between

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

135

them is expressed via a third variable. The third variable is called the parameter.
More
precisely, a relation expressed between two variables x and y in the form
x = f (t), y = g (t) is said to be parametric form with t as a parameter.
In order to find derivative of function in such form, we have by chain rule.

dy dx
dy
⋅
=
dx dt
dt

or

dy
dy
dx


≠ 0
= dt  whenever
dx 
dx
dt

dt

Thus

g ′(t )  dy
dx
dy

= f ′(t )  [provided f ′(t) ≠ 0]
=
 as = g ′(t ) and
f ′(t )  dt
dt

dx

Example 31 Find

dy
, if x = a cos θ, y = a sin θ.
dx

Solution Given that

x = a cos θ, y = a sin θ
Therefore

dx
dy
= – a sin θ,
= a cos θ
dθ
dθ

Hence

dy
dy
d θ = a cos θ = − cot θ
=
dx − a sin θ
dx
dθ

dy
, if x = at2, y = 2at.
dx
Solution Given that x = at2, y = 2at

Example 32 Find

So

dx
= 2at and
dt

Therefore

dy
dy
dt = 2a = 1
=
dx 2at t
dx
dt

Reprint 2024-25

dy
= 2a
dt
136

MATHEMATICS

Example 33 Find

dy
, if x = a (θ + sin θ), y = a (1 – cos θ).
dx

Solution We have

dx
dy
= a(1 + cos θ),
= a (sin θ)
dθ
dθ
dy
dy
d θ = a sin θ = tan θ
=
dx a (1 + cos θ)
dx
2
dθ

Therefore

dy

ANote It may be noted here that dx is expressed in terms of parameter only

without directly involving the main variables x and y.

Example 34 Find dy , if x 3 + y 3 = a 3 .
dx
Solution Let x = a cos3 θ, y = a sin3 θ. Then
2
2
2
2
3
3 + ( a sin 3 θ) 3
3
3
(
a
cos
θ
)
x +y =
2
2

= a 3 (cos θ + (sin θ) = a 3
2

Hence, x = a cos3 θ, y = a sin3 θ is parametric equation of x 3 + y 3 = a 3

Now

dx
dy
= – 3a cos2 θ sin θ and
= 3a sin2 θ cos θ
dθ
dθ

Therefore

dy
2
dy
d θ = 3a sin θ cos θ = − tan θ = − 3 y
=
dx − 3a cos 2 θ sin θ
x
dx
dθ

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

137

EXERCISE 5.6
If x and y are connected parametrically by the equations given in Exercises 1 to
10,
without eliminating the parameter, Find
1. x = 2at2, y = at4
3. x = sin t, y = cos 2t

dy
.
dx
2. x = a cos θ, y = b cos θ
4. x = 4t, y =

4
t

5. x = cos θ – cos 2θ, y = sin θ – sin 2θ

6. x = a (θ – sin θ), y = a (1 + cos θ) 7. x =

cos3 t
sin 3 t
, y=
cos 2t
cos 2t

t

x = a  cos t + log tan  y = a sin t 9. x = a sec θ, y = b tan θ

2
10. x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)

8.

−1

−1

11. If x = a sin t , y = a cos t , show that

dy
y
=−
dx
x

5.7 Second Order Derivative

Let

y = f (x). Then

dy
= f ′(x)
... (1)
dx
If f ′(x) is differentiable, we may differentiate (1) again w.r.t. x. Then, the
left hand
side becomes

d  dy 
  which is called the second order derivative of y w.r.t. x and
dx  dx 

d2y
. The second order derivative of f (x) is denoted by f ″(x). It is also
dx 2
denoted by D2 y or y″ or y2 if y = f (x). We remark that higher order derivatives
may be
defined similarly.
is denoted by

Reprint 2024-25
138

MATHEMATICS

d2y
, if y = x3 + tan x.
dx 2
Solution Given that y = x3 + tan x. Then
Example 35 Find

dy
= 3x2 + sec2 x
dx
Therefore

d2y
d ( 2
3x + sec2 x )
2 =
dx
dx
= 6x + 2 sec x . sec x tan x = 6x + 2 sec2 x tan x

Example 36 If y = A sin x + B cos x, then prove that

d2y
+ y=0.
dx 2

Solution We have

dy
= A cos x – B sin x
dx
and

Hence

d2y
d
(A cos x – B sin x)
2 =
dx
dx
= – A sin x – B cos x = – y
d2y
+y=0
dx 2

Example 37 If y = 3e2x + 2e3x, prove that

d2y
dy
− 5 + 6y = 0 .
2
dx
dx

Solution Given that y = 3e2x + 2e3x. Then

dy
= 6e2x + 6e3x = 6 (e2x + e3x)
dx
Therefore

Hence

d2y
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
dx 2
d2y
dy
−5
+ 6y = 6 (2e2x + 3e3x)
2
dx
dx
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

Example 38 If y = sin–1 x, show that (1 – x2)

d2y
dy
−x =0.
2
dx
dx

Solution We have y = sin–1 x. Then

dy
=
dx

1
(1 − x 2 )

dy
=1
dx

or

(1 − x 2 )

So

d 
dy 
2
 (1 − x ) .  = 0
dx 
dx 

or

(1 − x 2 ) ⋅

d 2 y dy d
+ ⋅
dx 2 dx dx

or

(1 − x 2 ) ⋅

d 2 y dy
2x
− ⋅
=0
2
dx 2 1 − x 2
dx

( (1 − x ) ) = 0
2
d2y
dy
−x =0
2
dx
dx
Alternatively, Given that y = sin–1 x, we have
Hence

(1 − x 2 )

y1 =

So
Hence

1
1 − x2

, i.e., (1 − x 2 ) y 2 = 1
1

(1 − x 2 ) . 2 y1 y2 + y12 (0 − 2 x) = 0
(1 – x2) y2 – xy1 = 0

EXERCISE 5.7
Find the second order derivatives of the functions given in Exercises 1 to 10.
1. x2 + 3x + 2

2. x 20

3. x . cos x

4. log x

5. x3 log x

6. ex sin 5x

7. e6x cos 3x
8. tan–1 x
9. log (log x)
10. sin (log x)
d2y
+ y=0
11. If y = 5 cos x – 3 sin x, prove that
dx 2

Reprint 2024-25

139
140

MATHEMATICS

12. If y = cos–1 x, Find

d2y
in terms of y alone.
dx 2

13. If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0

d2y
dy
− (m + n) + mny = 0
14. If y = Ae + Be , show that
2
dx
dx
mx

nx

15. If y = 500e7x + 600e– 7x, show that

16. If ey (x + 1) = 1, show that

d2y
= 49 y
dx 2

d 2 y  dy 
= 
dx 2  dx 

17. If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2

Miscellaneous Examples
Example 39 Differentiate w.r.t. x, the following function:
(i)

3x + 2 +

(ii) log7 (log x)

2x2 + 4

Solution
(i) Let y =

3x + 2 +

1
2

2 x2 + 4

= (3x + 2) 2 + (2 x + 4) 2

2
Note that this function is defined at all real numbers x > − . Therefore
3
1

−1 d
− −1 d
dy
1
 1
2
2
= (3 x + 2) 2 ⋅ (3 x + 2) +  −  (2 x + 4) 2 ⋅ (2 x + 4)
dx
2
dx
 2
dx
1

−
−
1
 1
2
= (3x + 2) 2 ⋅ (3) −   (2 x + 4) 2 ⋅ 4 x
2
2

3
−
2 3x + 2

2x
3

( 2x2 + 4) 2

2
This is defined for all real numbers x > − .
3
Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

141

log (log x)
(by change of base formula).
log 7
The function is defined for all real numbers x > 1. Therefore

(ii) Let y = log7 (log x) =

1 d
dy
(log (log x))
=
log
7 dx
dx
=

1
1
d
⋅ (log x)
log 7 log x dx

1
x log 7 log x

Example 40 Differentiate the following w.r.t. x.

(i) cos – 1 (sin x)

 sin x 
(ii) tan −1 

 1 + cos x 

 2 x +1 
(iii) sin −1 

 1 + 4x 

Solution
(i) Let f (x) = cos – 1 (sin x). Observe that this function is defined for all real
numbers.
We may rewrite this function as
f (x) = cos – 1 (sin x)

 π

= cos −1  cos  − x 


2


π
−x
2
Thus
f ′(x) = – 1.
sin
x 

(ii) Let f (x) = tan – 1 
 . Observe that this function is defined for all real
 1 + cos x 
numbers, where cos x ≠ – 1; i.e., at all odd multiplies of π. We may rewrite this
function as
=

−1  sin x 
f (x) = tan 

 1 + cos x 


 x
 x 
2 sin   cos   

2
2
−1

= tan 
x
2


2cos


2

Reprint 2024-25
142

MATHEMATICS

 x  x
−1 
= tan  tan    =
  2  2
 x
Observe that we could cancel cos   in both numerator and denominator as it
2

is not equal to zero. Thus f ′(x) =

1.
2

 2x + 1 
(iii) Let f (x) = sin – 1 
 . To find the domain of this function we need to find all
 1 + 4x 

x such that −1 ≤

2 x +1
≤ 1 . Since the quantity in the middle is always positive,
1 + 4x

we need to find all x such that

2 x +1
≤ 1 , i.e., all x such that 2x + 1 ≤ 1 + 4x. We
x
1+ 4

1
+ 2x which is true for all x. Hence the function
2x
is defined at every real number. By putting 2x = tan θ, this function may be
rewritten as

may rewrite this as 2 ≤

x +1

−1  2
f (x) = sin 
x
1 + 4 

x
−1  2 ⋅ 2 
sin
=

2
1 + ( 2 x ) 

−1  2 tan θ 
= sin 
1 + tan 2 θ 

= sin – 1 [sin 2θ]

= 2θ = 2 tan – 1 (2x)
Thus

f ′(x) = 2 ⋅

1
1 + (2 )

x 2

d x
(2 )
dx

2
⋅ (2 x )log 2
1 + 4x

2 x + 1 log 2
=
1 + 4x

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

143

Example 41 Find f ′(x) if f (x) = (sin x)sin x for all 0 < x < π.
Solution The function y = (sin x)sin x is defined for all positive real numbers.
Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)

1 dy
d
=
(sin x log (sin x))
y dx
dx

Then

= cos x log (sin x) + sin x .

1 d
⋅ (sin x)
sin x dx

= cos x log (sin x) + cos x

= (1 + log (sin x)) cos x
Thus

dy
= y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
dx

Example 42 For a positive constant a find

dy
, where
dx
a

1
t,

 1
and x =  t + 
 t
Solution Observe that both y and x are defined for all real t ≠ 0. Clearly
y=a

t+

( )

t+ d 
1
dy
d t +1
t
a
=
=
 t +  ⋅ log a
t
a
dt  t 
dt
dt

= a

Similarly

t+

1
1
t 1 −
log a

2 

dx
 1
= a t + 
dt
 t

t 

a −1

 1
= a t + 
 t

⋅
a −1

d  1
t + 
dt  t 

1

⋅ 1 − 2 
 t 

dx
≠ 0 only if t ≠ ± 1. Thus for t ≠ ± 1,
dt

Reprint 2024-25
144

MATHEMATICS

t+ 
1
dy
a t  1 − 2  log a
 t 
dy dt
=
=
a −1
dx dx
1
 1

a t +  ⋅  1 − 2 
dt
 t 
 t

t+

1
t log a

 1
at + 
 t
2
cos x
Example 43 Differentiate sin x w.r.t. e .

a −1

Solution Let u (x) = sin2 x and v (x) = e cos x. We want to find du = du / dx .

Clearly
dv dv / dx

du
dv
= 2 sin x cos x and
= e cos x (– sin x) = – (sin x) e cos x
dx
dx
2sin x cos x
2cos x
du
= − cos x
=
cos x
− sin x e
e
dv

Thus

Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. x the function in Exercises 1 to 11.
1. (3x2 – 9x + 5)9
2. sin3 x + cos6 x
3. (5x) 3 cos 2x
4. sin–1(x x ), 0 ≤ x ≤ 1
x
2 ,–2<x<2
2x + 7

cos −1

5.

6.
7.
8.
9.
10.

 1 + sin x + 1 − sin x 
π
cot −1 
,0<x<
2
 1 + sin x − 1 − sin x 
(log x)log x, x > 1
cos (a cos x + b sin x), for some constant a and b.
π
3π
(sin x – cos x) (sin x – cos x), < x <
4
4
xx + xa + ax + aa, for some fixed a > 0 and x > 0

Reprint 2024-25
CONTINUITY AND DIFFERENTIABILITY

11.

145

x2

x x −3 + ( x − 3) , for x > 3
2

π
π
dy
, if y = 12 (1 – cos t), x = 10 (t – sin t), − < t <
2
2
dx
dy
13. Find
, if y = sin–1 x + sin–1 1 − x 2 , 0 < x < 1
dx

12. Find

14. If x 1 + y + y 1 + x = 0 , for , – 1 < x < 1, prove that

dy
1
=−
dx
(1 + x )2
15. If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
3

  dy 2  2
1 +   
  dx  
d2y
dx 2
is a constant independent of a and b.
16. If cos y = x cos (a + y), with cos a ≠ ± 1, prove that

dy cos 2 ( a + y )
.
=
dx
sin a

17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find

d2y
.
dx 2

18. If f (x) = | x |3, show that f ″(x) exists for all real x and find it.
19. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the
differentiation,
obtain the sum formula for cosines.
20. Does there exist a function which is continuous everywhere but not
differentiable
at exactly two points? Justify your answer.

f ( x) g ( x ) h( x )
21. If y =

l
a

m
b

n
c

dy
=
, prove that
dx

f ′( x) g ′( x) h′( x)
l
a

m
b
2

n
c

22. If y = ea cos x , – 1 ≤ x ≤ 1, show that (1 − x 2 ) d y − x dy − a 2 y = 0 .

dx
dx 2
−1

Reprint 2024-25
146

MATHEMATICS

Summary

® A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point. A function
is continuous if it is continuous on the whole of its domain.

® Sum, difference, product and quotient of continuous functions are continuous.

i.e., if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g (x) is continuous.
(f . g) (x) = f (x) . g (x) is continuous.
f ( x)
 f 
 g  ( x) = g ( x) (wherever g (x) ≠ 0) is continuous.
 

® Every differentiable function is continuous, but the converse is not true.

® Chain rule is rule to differentiate composites of functions. If f = v o u, t = u
(x)
and if both

dt
dv
and
exist then
dx
dt

df dv dt
= ⋅
dx dt dx
® Following are some of the standard derivatives (in appropriate domains):
d ( −1 )
1
sin x =
dx
1 − x2

d ( −1 )
1
cos x = −
dx
1 − x2

1
d ( −1 )
tan x =
dx
1 + x2

d ( x) x
d
1
e =e
( log x ) =
dx
dx
x
® Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x). Here both f (x) and u (x) need to be positive for
this technique to make sense.

—v —

Reprint 2024-25