U YE MIN HTOO M.E [Y.T.

U] GRADE - 11 Mathematics

Chapter 11
Methods of Differentiation
In this chapter we shall consider real-valued functions whose domains are arbitrary subsets of R.
We first discuss the fundamental notion of limit of a function and the concept of derivatives. We
next introduce sum rule, different rule, product rule, quotient rule and chain rule that will be used
to calculate derivatives more efficiently. Finally, we shall explain the method of differentiation for
implicit functions.

11.1Limit of Functions
The basic idea of limit is to describe the behaviour of
functions when the independent variable approaches a
given value, For example, let us consider the behaviour
of the function
f (x) = 2x2
for x-values closer and closer to 2.
Here x would be 2.1, 2.01, 2.001, 2.0001, … from
the right of 2 as well as x would be 1.9, 1.99, 1.999, 1.9999, …
from the left of 2, It is seen that none of them equal to 2,
but x are selected closer to 2 on either left or right side of 2. Then we find the following;

x 1.9 1.99 1.999 1.9999 2 2.0001 2.001 2.01 2.1

f(x) 7.22 7.9202 7.992002 7.9920002 8.00080002 8.008002 8.0802 8.82

------------------------------------------------------  8  --------------------------------------------
Lest side Right side

It turns out that the values of f (x) get closer and closer to 8 from either side.
We write
lim 2x2 = 8 or 2 x2  8 as x  2.
x2

If the values of f (x) get closer and closer to a by taking the values of x
sufficiently close to c (but not equal to c), then we write
lim f (x) = a
x c

which is read as “ the limit of f (x) is a as x approaches c or a is a limit

of f” We can also write
f (x) → a as x → c

Notice the phrase “the value of x sufficiently close to c (but not equal to c)” in the definition of
limit. This means that in finding the limit of f(x) as x approaches c, we never consider x = c.

Limit of a Constant Function

For the constant function f (x) = a,
lim f (x) = lim a = a, where c is a real number.
xc x c

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Limit of Identity Function

For the identity function f (x) = x,
lim f (x) = lim x = c, where c is a real number.
xc x c

It is easily seen that

lim 5 = 5, lim (-2) = -2, lim x = 1, lim x = -2.
x2 x 0 x 1 x  -2

The limit laws

If lim f (x) and lim g (x) exist and c and k are real number, then
xc xc

1. Sum Rule; lim [f (x) + g (x)] = lim f (x) + lim g (x)

xc x c x c

2. Constant Multiple Rule; lim [k . f (x)] = k lim f (x)

x c x c

3. Product Rule; lim [f (x) . g (x)] = [ lim f (x)] [ lim g (x)]

xc xc x c

f (x) lim f (x)

x c
4. Quotient Rule; lim [ ] = , lim g (x)  0.
x c g(x) limg(x) x c
x c

5. Power Rule; lim [f (x)]n = [ lim f (x)]n , n is a positive integer

xc x c

6. Root Rule; lim n f (x) = n lim f (x) .

x c x c

(If n is even, we assume that lim f (x) 0.)

x c

Example 1.
Find (a) lim (x + 6) (b) lim (2  x) (c) lim (6x) (d) lim x( x  1).
x 3 x  -1 x 3 x  -1

Solution;
(a) lim (x + 6) = lim x + lim 6 = 3+ 6 = 9
x 3 x 3 x 3

(b) lim (2  x) = lim 2  lim x = 2 (1) = 3

x  -1 x  -1 x  -1

(c) lim (6x) = [ lim 6][ lim x] = 6 3 = 18

x 3 x 3 x 3

(d) lim x( x  1) = [ lim x][ lim x – 1] = (1) (2) = 2

x  -1 x  -1 x  -1

Example 2.
4x3 3x 1
Find lim .
x2 x 1

Solution;

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

4x3 3x 1 lim (4x3 3x- 1) 35

x 2
lim = = .
x2 x 1 lim (x 1) 3
x 2

Example 3.
4
Find (a) lim (2x  3)3 (b) lim 7x2 6 (c) lim (x3  2x + 1 ) 3 .
x2 x 0 x  -1

Solution;
(a) lim (2x  3)3 = [ lim (2x  3)]3 = 13 = 1
x2 x2

(b) lim 7x2 6 = lim (7x2  6) = 6

x0 x 0

4
(c) lim (x3  2x + 1 ) 3 = 3 lim (x3  2x 1)4
x  -1 x  -1

= 3 [lim (x3 2x 1)]4 = 3 24 = 23 2

x  -1

Limit of a Polynomial function

If P is a polynomial function, then

lim P (x) = P (c), where c is any number.
xc

Example 4.
Find lim (5x3  3x2 + 2x  1).
x2

Solution;
lim (5x3  3x2 + 2x  1) = 5  23  3  22 + 2  2  1 = 40  12 + 4  1 = 31
x2

When we consider the rational function’s behaviour near a particular point x0 that leads to division
by zero, the quotient rule in the Limit Laws cannot be used.

x2 9
Let us consider the function f (x) = .
x 3
This function is defined for all real number x except x = 3.

That is,
x2 9
f (x) = , x  3.
x 3
We can simplify that
x2 9 (x 3)(x 3)
f (x) = = = x + 3 for x  3.
x 3 x 3

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

The graph of f is the line y = x + 3 with the point (3, 6) removed which is shown as a “hole”.
Even though f (3) is not defined, we find the value off (x) as close as 6 by choosing x close
enough to 3.
x2 9
Thus lim = lim (x + 3) = 6.
x 3 x 3 x 3

Example 5.

x3  x2  4x  4 x2 3x x 11
Find (a) lim (b) lim (c) lim .
x 1 x 1 x 0 x x 0 x

Solution;

x3  x2  4x  4 x2 (x1) 4(x 1)

(a) lim = lim
x 1 x 1 x 1 x 1

(x1)(x2 4)
= lim
x 1 x 1
= lim (x  4) = 3.
2
x 1

x2 3x x(x 3)

(b) lim = lim = lim (x  3) = 3.
x 0 x x 0 x x 0

 
x 11  x 11 x 11
(c) lim . = lim 
x x 0  x 
x 0
 x 11
 

1 1
= lim = .
x 0
( x 11) 2

Exercise 11.1
1. Find the limits of the following;

x2 2x
(a) lim (x  3x) 3
(b) lim ( x  x  x)
2
(c) lim
x 2 x 4 x 5 x 2

x2  x
(d) lim (2x  5x + 1) 2
(e) lim (f) lim 4x 9
x  -3 x 1 x 1 x 0

1 4  x 1 9x x2
(g) lim  1 (h) lim (i) lim
x 0 x  x 4  x 1 x 1 x 9 3 x

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Solution;
(a) lim (x3 – 3x) = 23 - 3 (2) =8–6 = 2
x 2

(b) lim ( x 2  x - x) = lim x 2  x - lim x

x 4 x 4 x 4

= lim ( x 2  x) – 4 = 4 2  4 - 4 = 2 3 - 4
x 4

x2  2x x( x  2)
(c) lim = lim = lim x = 5
x 5 x2 x 5 x2 x 5

(d) lim (2x2 – 5x + 1) = 2 ( - 3 )2 - 5 ( - 3 ) + 1 = 18 + 15 + 1 = 34

x 3

x2  x x( x  1)
(e) lim = lim = lim x = 1
x 1 x 1 x 1 x 1 x 1

(f) lim 4x  9 = lim (4 x  9 ) = 4(0)  9 = 9 = 3

x 0 x 0

1 4 1 4 x4
(g) lim ( - 1) = lim ( . )
x 0 x x4 x 0 x x4
1 x
= lim ( . )
x 0 x x  4

1 1 1
= lim (- ) =- =-
x 0 x4 04 4

x 1 x 1 x 1
(h) lim = lim (  )
x 1 x 1 x 1 x 1 x 1
( x  1)( x  1)
= lim ( )
x 1 x 1
= lim ( x  1 ) = 1 + 1 = 2
x 1

9x  x2 9x  x 2 3 x
(i) lim = lim (  )
x 9 3 x x 9 3 x 3 x
x(9  x)(3  x )
= lim ( )
x 9 9 x
= lim x ( 3  x )
x 9

= ( lim x) ( lim ( 3  x ))
x 9 x 9

= 9 ( 3  9 ) = 54 .

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

One-Sided Limits of a Function

At the beginning of the chapter, we describe lim f (x) = a by saying that x gets closer to c on
x c

either left or right side of c. If we only approaches c from one side, we have a
one-sided limit. As x gets closer to c, but remains less than c, the corresponding value of f (x)
gets closer to L. The notation
lim f (x) = L
x c

is called the left- hand limits.

As x gets closer to c, but remaims greater than c, the corresponding value of f (x) get closer to R.
The notation
lim f (x) = R
x c

is called the right-hand limit.

Example 6.
Find lim |x| and lim |x|.
x 0 x 0

Solution;
x, if x  0
|x| =
- x, if x  0,
If x > 0, then |x| = x and x is getting close to 0,
lim |x| = lim x = 0.
x 0 x 0

If x  0, then |x| =  x and x is getting close to 0,

lim |x| = lim ( x) = 0.
x 0 x 0

Note In this example we see that lim |x| = lim |x| . Thus|x| has a limit 0 as x approaches 0.
x 0 x 0

Afunction f (x) has a limit as x approaches c if and only if left-hand and right-hand limits exist and
have the same value;
lim f (x) = a  lim f (x) = lim f (x) = a.
x c xc x c

However, some function show different behaviours on the two sides of an x-value. We often see
the following case
lim f (x)  lim f (x).
x c x c

In that case we say that f (x) has no limit when x closes to c.

Example 7.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

x2  x
Find the limits of f (x) = when x →0 if the limit of f (x) exists.
| x|

Solution;

x2  x
f (x) = , x  0.
| x|

x2  x
when x  0, f (x) = = x+1
x

x2  x
when x  0, f (x) = = x1.
x

x2  x
We observe that when x  0, lim = lim (x + 1) = 1 and
x 0 | x| x 0

x2  x
when x  0, lim = lim ( x  1) =  1 .
x 0 | x| x 0

x2  x x2  x
Hence lim  lim .
x 0 | x| x 0 | x|
Therefore f (x) has no limit when x closes to 0.

Infinite limits
1
We now consider the behaviour of function f (x) = for
x
values of x near 0. In the graph, the x-values are taken
closer and closer to 0 from the right side, the values of
f (x) are positive and increase larger and larger.
Similarly the x-values are taken closer and closer to 0
from the left side, the values of f (x) are negative and
decrease smaller and smaller.
More precisely,
when x =  1, f (x) =  1; when x = 1, f (x) = 1
when x =  0.1, f (x) =  10; when x = 0.1, f (x) = 10
when x =  0.01, f (x) =  100; when x = 0.01, f (x) = 100,etc.
1 1
Therefore lim =  . lim = .
x 0 x x 0 x

In writing these equations, we are not saying that the the limits exist because there are no real
1
number  and . So we say that the function f (x) = has no limits as x  0- and as
x
+
x 0 .

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Example 8.

1 1
Find the one-sided limits of (a) lim (b) lim .
x2 4 2x x2 4 2x

Solution;

1
The domain of function is x | x 2.
4 2x

1
(a) If x  2 and x is getting close to 2, the value of is negative and is getting smaller.
4 2x
That is

1
lim = .
x  2 4 2x

1
(b) If x  2 and x is getting close to 2, the value of is positive and is getting larger. That
4 2x
is

1
lim = .
x  2 4 2x

Continuous Functions
The graph of a function can be described as a continuous curve if it has no breaks or holes,
Therefore we now study what properties of a function can cause breaks or holes.

Continuity at a Point
Definition.
A function f is said to be continuity at a point x = c if the following conditions are satisfied.
1. f (c) is defined.
2. lim f (x) exists.
x c

3. lim f (x) = f (c).

x c

If one or more of the conditions of this definition fail to hold, then we will say that f is
discontinuous at x = c and c is a point of discontinuous of f.

Continuous Discontinuous at x = 0 Jump at x = 0

Fig. 11.1 Fig. 11.1 Fig. 11.1

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

In figure 11.3, the function f (x) has no limit as x  0 because its values jump at x = 0. For
positive value of x arbitrarily close to zero, f (x) = 1. For the negative value of x arbitrarily close
to zero, f (x) = 4. There is no single value approached by f (x) as x  0.

Continuity on an Interval
A function f is continuous at each number in an open interval (a, b). then we say that f is
continuous on (a, b). However when we consider the continuity of function on a closed interval
a, b or the end point of an interval of the form a, b) or (a, b, we need to consider the one-sided
limit at that endpoint. For example, the function graphed in the figure is continuous at the right
endpoint of the interval (a, b because
lim ( x ) = (b)
xb

but it is not continuous at the left endpoint because

lim ( x )  (a).
x a

In general, we say a function  is continuous from the left at c if

lim f (x) = f (c)
x c

and continuous from the right at c if

lim f (x) = f (c).
x c

Definition.
A function f is said to be continuous on a colsed interval
a, b if the following conditions are satisfied;
1. f is continuous on (a, b).
2. f is continuous form the right at a.
3. f is continuous form the left at b.
A continuous function is one that is continuous at every
point of its domain.

Example 9.
Show that a function f (x) = 4  x2 is continuous at every point of it domain 2, 2.

Solution;
If c is any point in the interval (2, 2), then
lim f (x) = lim 4  x2 = lim(4 x2 ) = 4 c2 = (c).
x c x c x c

Thus, f is continuous at each point is (2,2).

lim f (x) = lim  4  x2 = lim (4 x2 ) = 0 = f (2).
x  -2 x  -2 x -2

A function  is continuous from the left at 2.

lim f (x) = lim  4  x2 = lim (4 x2 ) = 0 = f (-2).
x  -2 x  -2 x -2

A function f is continuous from the right at 2. Hence, f is continuous on the closed interval 2,
2.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Example 10.
x2
The function y = - is continuous at every value of x
x 1
except x = 1. It has a point of discontinuity at x = 1 because
it is not defined there. Therefore the given rational function is
continuous on its domain ( , 1) and (1, ).

Note. Every polynomial function is continuous on R and every rational function is continuous on
its domain.

Limits at Infinity

If the values of variable x increase arbitrarily larger, then we write x → and if the values of
variable x decrease arbitrarily small, then we write x →  . Let us consider the limit of function f
1
(x) = for both x →  and x →  .
x

That is
1 1
x = 1, = 1; x = 1, = 1
x x
1 1
x = 10, = 0.1; x = 10, = 0.1
x x
1 1
x = 100, = 0.01; x = 100, = 0.01,etc.
x x
1 1
It turns out that lim = 0 and lim = 0. More generally, for n = 1,2....
x  x x - x

1 1
lim = 0, lim = 0.
x   xn x  -  xn

Example 11.

x2 x 1
Find; (a) lim (6x3  3x2  1) (b) lim (2x5 +3x2  x) (c) lim
x  x  - x  x3  2

Solution;
3 1
(a) lim (6x3  3x2  1) = lim [x3 (6   )] =
x  x  x x3

3 1
(b) lim (2x5 +3x2  x) = lim [x5 (2 + 3
 )] = 
x  - x  - x x4

(c) We first divide the numerator and denominator by the highest power of x in the denominator.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1 1 1
x2 x 1  
lim = lim x x2 x3 = 00 0 = 0
x  x3  2 x  2 10
1 3
x

Exercise 11.2
1. Find the one-sided limit;
1 1 1 1
(a) lim (b) lim (c) lim (d) lim
x  1 1 x x 1 1 x x2 x2 x2 x2

x 2 x2 x x
(e) lim (f) lim (g) lim (h) lim
x2 x2  4 x2 x2  4 x 5 x2 5x x 5 x2  5x

Solution;
1
(a) lim
x  1 1 x

1
The domain of function is {x | x  1},
1 x
1
If x > 1 and x is getting close to 1, the value of is negative is getting smaller.
1 x
1
That is lim =-
x 1 1 x

1
(b) lim
x 1 1 x
1
The domain of function is {x | x  1},
1 x
1
If x < 1 and x is getting close to 1, the value of is positive is getting larger.
1 x
1
That is lim = 
x 1 1 x

1
(c) lim
x2 x2
1
The domain of function is {x | x  2},
x 2
1
If x > 2 and x is getting close to 2, the value of is positive is getting larger.
x 2
1
That is lim = 
x2 x 2

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1
(d) lim
x2 x 2
1
The domain of function is {x | x  2},
x 2
1
If x < 2 and x is getting close to 2, the value of is negative is getting smaller.
x 2
1
That is lim =-
x2 x 2

x2 x2 1
(e) lim = lim = lim
x2 x2  4 x2 ( x  2)( x  2) x2 x2
1
The domain of function is {x | x  2},
x 2
1
If x > 2 and x is getting close to 2, the value of is positive is getting larger.
x 2

x 2 1
That is lim = lim = 
x2 x 4 x  2 x 2
2

x2 x2 1
(f) lim = lim = lim
x2 2
x 4 x2 ( x  2)( x  2) x2 x 2
1
The domain of function is {x | x  2},
x 2
1
If x < 2 and x is getting close to 2, the value of is negative is getting smaller.
x 2

x 2 1
That is lim = lim = -
x2 2
x 4 x  2 x 2

x x 1
(g) lim = lim = lim
x 5 2
x 5x x 5 x(x 5) x  5 x 5
1
The domain of function is {x | x  5},
x 5
1
If x > 5 and x is getting close to 5, the value of is positive is getting larger.
x 5

x 1
That is lim = lim = 
x 5 x2  5x x  5 x 5

x x 1
(h) lim = lim = lim
x 5 x2 5x x  5 x(x 5) x 5 x 5
1
The domain of function is {x | x  5},
x 5

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1
If x < 5 and x is getting close to 5, the value of is negative is getting smaller.
x 5

x 1
That is lim = lim = -
x 5 2
x  5x x  5 x 5

2. Show that the limit do not exist for the following;

|x| x 2  2x x2 1
(a) lim (b) lim (c) lim
x 0 x x 0 |x| x 1 | x 1|

Solution;
| x|
(a) Let f (x) = , x  0
x
x
When x > 0, f (x) = = 1
x
x
When x < 0, f (x) = = -1
x
|x|
when x > 0 , lim = lim 1 =1
x 0x x 0

| x|
when x < 0 , lim = lim (-1) =-1
x 0 x x 0

lim | x |  lim | x |
x 0 x x 0 x
f (x) has no limit when x close to 0.

x 2  2x
(b) Let f (x) = , x 0
|x|
x2  2x
when x > 0, f (x) = = x–2
x
x 2  2x
when x < 0, f (x) = = -x+2
x
x 2  2x
when x > 0 , lim = lim (x – 2) = - 2
x 0 |x| x 0

x 2  2x
when x < 0 , lim = lim (- x + 2) = 2
x 0 |x| x 0

x 2  2x x 2  2x
lim  lim
x 0 | x| x 0 |x|
f (x) has no limit when x close to 0.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

x 2 1
(c) Let f (x) = , x 1
| x 1|
x2  1 ( x  1)( x  1)
when x > 1, f (x) = = =x+1
x 1 x 1
x2 1 ( x  1)( x  1)
when x < 1, f (x) = = = - ( x + 1)
 ( x  1)  ( x  1)
x2 1
when x > 1 , lim = lim (x +1) =2
x 1 | x 1| x  1

x2 1
when x < 1 , lim = lim (- x – 1) = - 2
x 1 | x 1| x  1

x 2 1 x2 1
lim  lim
x 1 | x 1| x 1 | x 1|
f (x) has no limit when x close to 1.

3. Find the limit of the following functions;

3x 5 3x2  7
(a) lim (b) lim (c) lim (1 3x 4x7)
x  12x 1 x  x3  2 x -

2
x  x 10 2
x 2 2
(d) lim (e) lim (f) lim (1 - )
x  x 1 x  2x 1 x  - x

Solution;
5
3 30 1
3x  5 x
(a) lim = lim = =
x   12 x  1 x 
12 
1 12  0 4
x

3 7
 3
3x 2  7 00 0
(b) lim = lim x x = = = 0
x   x3  2 x  2 1 0 1
1 3
x

1 3
(c) lim (1 – 3x – 4x2) = lim [ x 7 (7
 6  4) ] = 
x  x
x  x
10
x 1 
x 2  x  10 x =  1 0 = 
(d) lim = lim
x  x 1 x  1 1 0
1
x
2 2
x 2 (1  ) 1
x2  2 x2 x2 1 0 1
(e) lim = = = =
x  2x 1 2 1 20 2
x (2  ) 2
x x

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

2
(f) lim (1 - ) = 1–0=1
x - x

4. Determine where  is continuous.

x 3
(a) f (x) = (b) f (x) = x (c) f (x) = x2 + 3x - 1
2x 1

Solution;
x3 1
(a) f (x) = , x -
2x  1 2
1
Domain of f = {x | x  R, x  - }
2
1 1
 f is continuous on its domain (-  , - ) and (- ,  ).
2 2

(b) f (x) = x , x  0
Domain of f = {x | x  0}
 f is continuous on its domain [0,  ).

(c) f (x) = x2 + 3x - 1
f is a polynomial function.
 f is continuous on R.

5. Find the value of x at which  is not continuous.

x2  2x 1 x 1 x2  4
(a) ( x ) = (b) ( x ) = + 2 (c) ( x ) =
x2 x x 1 | x  2|

Solution;
x2  2x
(a) ( x ) = , x - 2
x 2
f is not defined at x = -2.
 f is not continuous at x = -2.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1 x 1
(b) ( x ) = + 2 , x  0, x   1
x x 1
f is not defined at x = 0, x = 1 and x = -1.
 f is not continuous at x = 0, x = 1 and x = -1.

x2  4
(c) ( x ) = , x 2
| x  2|
|x - 2| is a modulus function, f has a jumpat x = 2.
 f is not continuous at x = 2.

11.2 Derivatives
let us consider the point P ( x0 , ( x0 )) on the graph of the function y = ( x ) and
Q ( x0 + h, ( x0 + h)) is another point on the graph . The slope of the curve at the point P is
f (x0  h)  f (x0 )
m = lim
h  0 h
if the limit exists, The tangent line to the curve at P is the line through P with this slope.

Example 12.
Find the slope of the curve y = x2 at any point x = a. What are the slope at the point x = 2 and x =
 1? Where does the slope equal 1?

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Solution;
Here f (x) = x2. The slope at (a, a2) is
f (a h)  f (a ) (a  h)2  a
2
lim = lim
h  0 h h  0 h
a2  2ah h2  a2
= lim
h  0 h
= lim (2a + h) = 2a.
h  0

Thus the slope at the point x = a is 2a. When a = 2, the slope is 4 and when a =  1, the slope is 
1
2. If the slope is 1, then 2a = 1. Thus a = . Therefore the curve has slope 1 at the point
2
1 1
( ).
2, 4

Derivative at a Point
f (x0  h)  f (x0 )
The expression , h  0 is called the Difference quotient of  at x0 with an
h
increment h. If the difference quotient has a limit as h approaches zero, it is given a special name
and notation.

Definition.
The derivative of a function f at a point x0, denoted by f  ( x0) is
(x0  h)  (x0)
f  ( x0 ) = lim
h  0 h
if this limit exists.

Example 13.
Find the derivative of  ( x ) = x2 + 5 at 3. That is , find f  (3).

Solution;
Since (3) = 32 + 5 = 14, we have
f (3h)  f (3) (3 h )2 514 h2 6h
= =
h h h
The derivative of  at 3 is
f (3h)  f (3) h(h6)
f  (3) = lim = lim = lim (h + 6) = 6.
h  0 h h  0 h h  0

Example 14.
Find the derivative of  ( x ) = x3 at c. That is , find f  (c).

Solution;
Since (c) = c3 , we have

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

f (c h)  f (c) (c  h )3 c3 3c2h3ch2 h3

= =
h h h
The derivative of  at c is
f (c h)  f (c) h(3c2 3ch h2)
f  (c) = lim = lim = lim (3c2  3ch h2) = 3c2 .
h  0 h h  0 h h  0

The Derivative as a Function

We now discuss the derivative as a function derived from  by considering the limit at each point
x in the domain of .

Definition.
The function f  defined by the formula
f (x h)  f (x)
f  ( x) = lim
h  0 h

is called the derivative of  with respect to x. The domain of f  consits of all x in the domain
of  for which the limit exists.

There are many ways to denote the derivative of a function y = ( x ). Some common notations
for the derivative are
dy df d
f  ( x ) = y = = =  ( x ).
dx dx dx
The process of calculating a derivative is called differentiation.

Example 15
Differentiate x2 + 3x + 6 with respective to x by using the definition.

solution;
Let ( x ) = x2 + 3x + 6
( x + h) = ( x + h)2 + 3 ( x + h) + 6 = x2 + 2xh + h2 + 3x + 3h + 6
= (x2 + 3x + 6) + (2xh + h2 + 3h)
Thus
f (x h)  f (x) 2xhh2 3h
f  ( x ) = lim = lim = lim ( 2x + h + 3) = 2x + 3.
h  0 h h  0 h h  0

Example 16.
1
Differentiate of  ( x ) = with respective to x by using the definition.
x

solution;
1 1
Since  ( x ) = , we have ( x + h) =
x x h
Then

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1 1 x (x  h) h
( x + h)  ( x ) =  = =
x h x (x  h)x (x  h)x
Therefore
df f (x h)  f (x) 1 1
= lim = lim =
dx h  0 h h  0 (x  h)x x2

Example 17.
Find the derivarive of  ( x ) = x for x  0 by using the definition.

solution;
f (x h)  f (x) xh  x
f  ( x ) = lim = lim
h  0 h h  0 h
( x  h  x)( x h  x) h 1
= lim = lim = .
h  0 h( x  h  x) h  0 h( x  h  x) 2 x

Note that some functions may fail to have the derivative. Thus we need to study the right-hand
and left-hand limit of difference quotient of a function at any point of the function’s domain.

Example 18.
Show that the function y = |x| has no derivative at x = 0.

Solution;
f (0h)  f (0) | h| h
To the right of the orgin, lim = = = 1.
x 0 h h h
f (0h)  f (0) | h| h
To the left, lim = = = 1.
x  0 h h h
We observe that
f (0h)  f (0) f (0h)  f (0)
lim  lim .
x 0 h x 0 h
Hence there is no dervative at the origin.

Exercise 11.3
1. By using the definition, find the dervative of each function at the given number.
(a) f (x) = x3 – x + 1 at 3 (b) f (x) = x2  2x at 0
(c) f (x) = - 5x + 1 at 2 (d) f (x) = x  2 at 4

Solution;
(a) f (x) = x3 – x + 1 at 3
f (3) = 33 - 3 + 1 = 25
f (3 + h) = (3 + h)3 - (3 + h) + 1
= 33 + 3 (3)2 h + 3 (3) h2 + h3 - 3 - h + 1
= 27 + 27 h + 9 h2 + h3 - 3 - h + 1

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

= h3 + 9 h2 + 26h + 25
f (3 + h) – f (3) = h3 + 9 h2 + 26h + 25 – 25
= h3 + 9 h2 + 26h
f (3h)  f (3) h3 +9 h2 + 26h
= = h2 + 9h + 26
h h
f (3h)  f (3)
f  (3) = lim
h  0 h
= lim (h2 + 9h + 26) = 26.
h  0

(b) f (x) = x2 – 2x at 0
f (0) = 02 – 2(0) =0
f (0 + h) = (0 + h)2 - 2 (0 + h)
= h2 - 2h
f (0 + h) – f (0) = h2 - 2h – 0 = h2 - 2h
f (0h)  f (0) h2 - 2h
= = h-2
h h
f (0h)  f (0)
f  (0) = lim
h  0 h
= lim (h - 2) = - 2.
h  0

(c) f (x) = - 5x + 1 at 2
f (- 2) = - 5 (-2) + 1 = 11
f (- 2 + h) = - 5 (-2 + h) + 1
= 10 – 5h + 1 = - 5h + 11
f (- 2 + h) – f (- 2) = - 5h + 11 - 11 = - 5h
f (2h)  f (2) - 5h
= = -5
h h
f (2h)  f (2)
f  (3) = lim
h  0 h
= lim (-5) = -5.
h  0

(d) f (x) = x  2 at 4
f (4) = 42 = 6
f (4 + h) = (4h)  2 = h6
f (4 + h) – f (4) = h 6 - 6
f (4h)  f (4) h 6 - 6 h 6 - 6 h 6  6
= =
h h h h 6  6
h  6- 6 h 1
= = =
h( h6  6) h( h6  6) h 6  6
f (4h)  f (4)
f  (4) = lim
h  0 h

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1 1 1 6
= lim = = =
h  0 h 6  6 6 6 2 6 12

2. Find the dervatives of each function with respective to x.

1
(a) f (x) = x2 – 2x + 1 (b) f (x) = 2
x

Solution;
(a) f (x) = x2 – 2x + 1
f (x + h) = (x + h)2 – 2 (x + h) + 1
= x2 + 2hx + h2 – 2x - 2h + 1
f (x + h) – f (x) = x2 + 2hx + h2 – 2x - 2h + 1 – (x2 – 2x + 1)
= x2 + 2hx + h2 – 2x - 2h + 1 – x2 + 2x - 1
= 2hx + h2 - 2h
f (x + h) - f (x) 2hx + h2 - 2 h
= = 2x + h - 2
h h
f (x + h) - f (x)
f  (x) = lim
h  0 h
= lim (2x + h - 2) = 2x - 2.
h  0

1
(b) f (x) =
x2
1
f (x + h) =
( x  h) 2
1 1 x 2  ( x  h) 2
f (x + h) – f (x) = 2
- 2 =
( x  h) x x 2 ( x  h) 2
x 2  x 2  2 hx  h 2  2 hx  h 2 h ( 2 x  h )
= = =
x 2 ( x  h) 2 x 2 ( x  h) 2 x 2 ( x  h) 2
f (x + h) - f (x) h ( 2 x  h ) 1  2x  h
= =
h x 2 ( x  h) 2 h x 2 ( x  h) 2
f (x + h) - f (x)
f  (x) = lim
h  0 h
 2x  h  2x 2
= lim ( ) = =- .
h  0 x 2 ( x  h) 2 x 2 ( x) 2 x3

3. Show that the function y = |x – 5| has no derivative at x = 5.

Solution;
Let f (x) = |x – 5|
f (5) = |5 – 5| = 0
f (5 + h) = |5 + h – 5| = |h|
f (5 + h) - f (5) = |h| - 0 = |h|
f (5h)  f (5) | h|
=
h h

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

f (5h)  f (5) | h| h
x > 5, lim = = = 1
x 5 h h h

f (0h)  f (0) | h| h
x < 5, lim = = = 1
x 5 h h h
f (5h)  f (5) f (5h)  f (5)
lim  lim .
x 5 h x 5 h
Hence there is no dervative at x = 5.

11.3 Differentiation Rules

In this section, we introduce the differentiation formulas of constant functions and power
functions. Then you will study the derivative of sum rule, product rule, quotient rule and chain
rule.

Derivative of a constant function

df d
If f (x) = c, c is a constant function, then = c = 0.
dx dx

At every value of x, we find that

f ( x  h)  f ( x)
f ' (x) = lim
h0 h
cc
= lim = lim 0 = 0
h0 h h 0

Power Rule
d n
If n is a positive integer, then x = n xn – 1.
dx

Proof
Let f (x) = xn where n is a positive integer.
f (x + h) = (x + h)n
n ( n  1) n – 2 2
= xn + n.xn – 1 . h+ .x .h + … + hn.
2
n ( n  1) n 2
= xn + h [ n . xn – 1 + .x .h+…]
2
n ( n  1) n – 2
 f (x + h) – f (x) = h [ n. xn  1 + .x . h + …]
2

d f ( x  h)  f ( x)
f (x) = lim
dx h0 h

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

n(n  1) n  2
h[n.x n 1  .x .h  ... 
= lim 2
h0 h
n(n  1) n 2
= lim [n.x n 1  .x .h  ...] = n. xn – 1
h0 2

Power Rule (General Version)

d n
If n is any rational number, then x = n xn – 1 for any x where the power
dx
xn and xn-1 are defined.

Example 19.
Differentiate the following with respect to x.
2 4
1
(a) x4 (b) x3 (c) x 3 (d)
x3

Solution;
d 4
(a) (x ) = 4x41 = 4x3
dx
2 2 1
d 3 3 31 3 3
(b) (x ) = x = x
dx 2 2
4 7
d 4 4 3 1 4 3
(c) ( x 3) =  x =  x
dx 3 3
d 1 d 3
(d) ( ) = (x ) = 3x31 = 3x4
dx x3 dx

Derivative of Constant Multiple Rule

d d
If f is a differentiable function of x and c is a constant, then (c f (x)) = c f (x).
dx dx

Proof
d cf ( x  h)  cf ( x) f ( x  h)  f ( x) d
(c f (x)) = lim = c lim = c f (x)
dx h  0 h h  0 h dx

Example 20.
dy
Find for (a) y = 3x2 (b) y =  x (c) y = 2 x .
dx

Solution;

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

dy d 2
(a) = ( 3x2 ) = 3 (x ) = 3  2x = 6x
dx dx

dy d dy
(b) = ( x) = ( 1.x) = 1  ( x) = 1
dx dx dx
(26)
1  1 1
dy d 2 1  2 1 
(c) = ( 2 x ) = 2 (x ) = 2  x = x 2
dx dx 2

For the curve y = f (x), the slope of the tangent

dy
 1 at the point (x1, y1) is the value of at x = x1,
dx
hence, the equation of the tangent at (x1, y1) is

y – y1 = f ' (x1) (x – x1).

The line  2 which is perpendicular to the tangent

 1 at (x1, y1) is called the normal to the curve at
1
(x1, y1). Hence its slope is the value of -
f ' (x 1 )
where f ' (x1)  0. The equation of the normal at (x1, y1) is

1
y – y1 = - (x – x1).
f ' (x 1 )

Example 21.
Find equations of the tangent line and normal line to the curve  (x) = x2 at the point (1, 1).

Solution;
 (x) = x2, then f  (x) = 2x .
At (1, 1), 𝑓 (𝑥 ) = f  (1) = 2.
The equation of the tangent line at (1, 1) is
𝑦 − 𝑦 = 𝑓′(𝑥 )(𝑥 − 𝑥 )
y  1 = 2 (x – 1)
y = 2x - 1.
1
The slope of normal line is - .
2
Thus the equation of the tangent line at (1, 1) is
𝑦−𝑦 = ( )
(𝑥 − 𝑥 )
1
y1 = - (x – 1)
2
1 3
y = - x+ .
2 2

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Sum Rule

If f and g are differentiable at x, them the sum f + g is differentiable at x,

d d d
(f (x) + g (x)) = f (x) + g (x)
dx dx dx
(or)
(f + g) ' (x) = f ' (x) + g ' (x)

Proof
d
( + g) ' ( x ) = ( + g) ( x )
dx
d
= (f (x) + g (x))
dx

= lim
 f (x h)  g(x h) f (x)  g(x)
h0 h
f (x  h)  f (x) g(x  h)  g(x) 
= lim   
h0  h h 
f (x h)  f (x) g(x h)  g(x)
(+g) ' ( x ) = lim lim
h0 h h0 h
d d
= ( x ) + g( x )
dx dx
= ' ( x ) + g' ( x ).
Combining the Sum Rule and the Constant Multiple Rule, we get
d d d
(( x )  g( x )) = ( x ) + (g( x ))
dx dx dx
d d
= ( x ) + (1) g( x )
dx dx
d d
= ( x )  g( x )
dx dx

Example 22.
Find the derivative of y = x3 + 2x2 – 3x – 6.

Solution;
dy d
= ( x3 + 2x2 – 3x – 6 )
dx dx
dx 3 dx 2 dx d
= +2 -3 - 6
dx dx dx dx
= 3x2 + 2 (2x) – 3 (1) – 0
= 3x2 + 4x – 3

25
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Product Rule

If  and g are differentiable at x,then the product g is differentiable at x,

d d d
(( x )g( x )) = ( x ) g( x ) + g( x ) ( x )
dx dx dx
or
( g) ' ( x ) = ( x ) g ' ( x ) + g( x )  ' ( x ).

Proof
d f (x  h)g(x h)  f (x)g(x)
(( x ) g( x )) = lim
dx h  0 h
f (x h)g(x h)  f (x  h)g(x)  f (x  h)g(x)  f (x)g(x)
= lim
h  0 h

= lim
 f (x h)(g(x h)  g(x)) g(x)(f (x h)  f (x))
h  0 h
 g(x  h)  g(x) f (x  h)  f (x) 
= lim  f (x  h).  g(x). 
h  0
 h h 
d d
= lim ( x  h ) g( x ) + g( x ) ( x )
h  0 dx dx

Since lim ( x  h ) = ( x ), we get

h  0

d d d
(( x )g( x )) = ( x ) g( x ) + g( x ) ( x ).
dx dx dx

You may notice that while the derivative of the sum of two differentiable functions is the sum of
their derivatives, the derivative of the product of two differentiable functions is not the product of
their derivatives.

Example 23.
Differentiate ( x 3  2 x ) (3 x 4  2) with respect to x .
Solution;
d
dx
 
( x 3  2 x )(3 x 4  2) = ( x 3  2 x )
d
dx
(3 x 4  2) + (3 x 4  2)
d 3
dx
( x  2 x)

= ( x 3  2 x ) (12 x 3 ) + (3 x 4  2) (3 x 2  2)
= 21x 6  30x 4  6 x 2  4

26
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Quotient Rule
f (x)
If  and g are differentiable at x and if g (x)  0, then the quotient is differentiable at x,
g(x)

d d
g(x) f (x)  f (x) g(x)
d  f (x)  dx dx
  =
dx  g(x)  (g(x))2
(or)

f g(x) f (x)  f (x)g(x)
  ( x )=
g (g(x))2

Proof
d  1  g(x)
First we proof   = - .
dx  g(x)  (g(x))2
 1 1  g(x)  g(x  h)
  
d  1 
  = lim g(x  h) g(x)  = lim
g(x  h)g(x)
dx  g(x)  h0 h h  0 h
1 g(x)  g(x  h)
= lim .
h  0 h g(x  h)g(x)
 g(x  h)  g(x) 1 
= lim   
h  0
 h g(x  h)g(x)
d 1
= g(x) lim .
dx h  0 g(x  h)g(x)
Since lim g(x  h) = g(x) , we get
h  0

d  1  g(x)
  =  .
dx  g(x)  (g(x))2

Next we use the product rule to obtain


d  f (x)  d  1  1  1 
  =  f (x)  =  ' ( x) . ( x )  
dx  g(x)  dx  g(x)  g(x) +  g(x) 
f ( x) f (x)g(x)
= 
g( x) g(x)2
g(x) f (x)  f (x)g(x)
= .
(g(x))2

Example 24.
2x 1 1 x 2
Differentiate (a) , (b) and (c) with respect to x.
5x 2 3x 4 2
x  3x 1

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Solution;
d d
(5x  2) (2x 1)  (2x 1) (5x  2)
d 2x 1 dx dx
(a) =
dx 5x 2 (5x  2)2
(5x 2)2 (2x 1)5 9
= =
(5x 2) 2
(5x  2)2

d
(3x  4)  0  (3x  4)
d 1 dx 3
(b) = =
dx 3x  4 (3x  4) 2
(3x  4)2 `

d d
(x2  3x 1) (x  2)  (x  2) (x2  3x 1)
d x2 dx dx
(c) =
dx x2  3x 1 (x2  3x 1)2
(x2  3x 1)  (x  2)(2x  3)
=
(x2  3x 1)2
(x2  3x 1)  (2x2  x  6)
=
(x2  3x 1)2
 x2  4x  5
=
(x2  3x 1)2

When we consider the differentiation of the function

F (x) = x2  x ,
we cannot calculate F (x) directly.

Observe that F is a composite function, that is if we let y = ( u ) = u and if

u = g ( x ) = x 2  x , we can write y = F (x) = (g( x )) = (  g) ( x ). We have studied how to
differentiate both  and g, so we can find the derivative of F =   g in terms of the derivatives of
 and g. This is called the Chain Rule.

The Chain Rule

If y =  ( u ) is differentiable at u and u = g ( x ) is differentiable at x, then the function
y =  (g( x )) is differentiable at x, and
dy dy du
= . .
dx du dx

Since y = (g( x )) = (  g) ( x ),
(  g) ( x ) =  (g( x )) . g( x ) .

Example 25. Find y( x ) is y( x ) = x2  x

28
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Solution;
Let u = x 2  x . Then y = u .
1
dy du x
1 1 2 .
y( x ) = . = (2 x 1) = (2 x 1) =
du dx 2 u 2
2 x x
2
x x

The Power Rule Combined with the Chain Rule

If n is any real number and u = g( x ) is differentiable, then
d n
dx
 
u = nun1
du
dx
or
d
dx
( g ( x )) n = n ( g ( x )) n1 . g( x ) .

Example 26.
Differentiate y = (x 2 1)5 with respect to x .

Solution;
y = (x2 1)5
dy d 2 5 d 2
= (x 1) = 5(x2 1)4 (x 1) = 5(x2 1)4 . 2 x = 10 x (x2 1)4 .
dx dx dx

Example 27.

3
 x 1
Find Find f ( x ) if f ( x ) = 
 x 1
Solution;
3
 x 1
f ( x) =  
 x 1
2
 x 1 d  x 1
f ( x ) = 3    
 x 1 dx  x 1
2
 x 1 (x 1).1(x 1).1 6(x 1)2
= 3  =
 x 1 (x 1)2 (x 1)4

Example 28.
Find f ( x ) if f ( x )= (x2  2x)3 (x4 1)2 .

Solution;
f ( x) = (x2  2x)3 (x4 1)2
d 2 d
f ( x ) = [ ( x  2 x) 3 ] (x4 1)2 + (x2  2x)3 [ ( x 4  1) 2 ]
dx dx

29
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

d d
= [ 3( x 2  2 x) 2 (2 x 2  2 x) ] (x4 1)2 + (x2  2x)3 [ 2( x 4  1) ( x 4  1) ]
dx dx
= 3 (x2  2x)2 (4x  2) (x4 1)2 + 2 (x2  2x)3 (x4 1) (4x3) .

Higher order derivatives

If y = f (x) is a differentiable function, then its derivative f '(x) is also a function of x. If f ' is also
differentiable, then we can differentiate f ' to get a new function of x denoted by
f ". The function f " is called the second derivative of f. It can be written as
df ' d2 f d 2 y d dy dy '
f "(x) = = = = ( )= = y"
dx dx 2 dx 2 dx dx dx
If y = f (x), then successive derivatives can be denoted by
f ', f " , f "' , f(4), f(5), ...
For example, if f (x) = 3x4, then f '(x) = 12x3 and we have
d d
f "(x) = f '(x) = (12x3) = 36x2 .
dx dx
d d
f "' (x) = f " (x) = (36x2) = 72x .
dx dx

Example 29.
Find y and y'' for y = (3x2  2x 1)2 .

Solution;
y = (3x2  2x 1)2
y( x ) = 2 (3x2  2x 1) (6x  2)
d d
y ( x ) = 2 (6x  2) (3x2  2x 1) + 2 (3x2  2x 1) (6x  2)
dx dx
= 2 (6x  2) (6x  2) + 12 (3x2  2x 1) = 4 (27x2 18x  5)

Exercise 11. 4
1. Differentiate the following with respect to x.
x3 1
(a) x4 – 5x3 + 2x -1 (b) (c) 2x +
x 1 2x
1
(d) (2x2  2) (x2  3x)2 (e) ( 3 + x2) 3  x2 (f)
x 1
3x  5 2x  7 x2 1
(g) (h) (i)
2x 2  7 x7 x 2 1

Solution;
d d d d d
(a) (x4 – 5x3 + 2x -1) = (x4 ) – (5x3 ) + (2x) – (1)
dx dx dx dx dx

30
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

= 4x3 - 15x2 – 2

d d
( x  1) ( x  3 )  x  3 ( x  1)
d x3 dx dx
(b) =
dx x  1 x  12
1
1 
( x  1) ( x  3) 2  x  3 (1)
= 2
x  12
x 1
 x3
2 x3 x  1  2x  6 1  x5
= = =
x  12 2 x3 x  12
2 x  3 ( x  1) 2

d 1 d d 1 -1
(c) (2x + ) = (2x) + ( x )
dx 2x dx dx 2
1 1
= 2 + (-1) (x)-2 = 2-
2 2x 2
d d d
(d) (2x2 + 2) (x2 - 3x)2 = (2x2 + 2) (x2 - 3x)2 + (x2 - 3x)2 (2x2 + 2)
dx dx dx
d
= (2x2 + 2) 2 (x2 - 3x) (x2 - 3x) + (x2 - 3x)2 (4x)
dx
= (2x2 + 2) 2 (x2 - 3x) (2x2 - 3) + 4x (x2 - 3x)2
= 2 (2x2 + 2) (x2 - 3x) (2x2 - 3) + 4x (x2 - 3x)2

d d d
(e) ( 3 + x2) 3  x2 = ( 3 + x2) 3  x2 + 3  x2 ( 3 + x2)
dx dx dx
1
1  d
= (3 + x2) (3  x 2 ) 2 (3 - x2) + 3  x 2 (2x)
2 dx
1
1 
= (3 + x2) (3  x 2 ) 2 (- 2x) + 2x 3  x 2
2
 3x  x 3
= + 2x 3  x 2
3 x 2

 3x  x 3  6 x  2 x 3
=
3  x2
 3 x 3  3x
=
3  x2

d d
d ( x  1) (1)  (1) ( x  1)
1 dx dx
(f) =
dx ( x  1) ( x  1) 2
( x  1)(0)  (1)(1)
=
( x  1) 2

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

1
= -
( x  1) 2
d d
( 2 x 2  7) (3 x  5)  (3x  5) (2 x 2  7)
d 3x  5 dx dx
(g) =
dx 2 x 2  7 ( 2 x 2  7) 2
( 2 x 2  7 )3  (3 x  5) 4 x
=
(2 x 2  7) 2
6 x 2  21  12 x 2  20 x
=
( 2 x 2  7) 2
21  20 x  6 x 2
=
(2 x 2  7) 2

d d
x7 (2 x  7)  (2 x  7) x7
d 2x  7 dx dx
(h) =
dx x  7 ( x  7 )2
1
1 
x  7 (2)  (2 x  7) ( x  7) 2
= 2
x7
2x  7
2 x7 
2 x7
=
x7
4 x  28  2 x  7 1
=
2 x7 x7
2 x  35
=
2(x  7) x  7

d x2 1 d x2 1
(i) =
dx x 2 1 dx x2 1
d d
x2 1 x2 1  x2 1 x2 1
= dx dx
( x 2  1) 2
1 1
 d 2  d 2
x 2  1 12 ( x 2  1) 2 ( x  1)  x 2  1 12 ( x 2  1) 2 ( x  1)
= dx dx
x2 1
1 1
 
x 2  1 12 ( x 2  1) 2 ( 2 x )  x 2  1 12 ( x 2  1) 2 (2 x)
=
x2 1

x x2 1 x x2 1

= x2 1 x2 1
2
x 1
x3  x  x3  x
1
= 2
2
x 1 x 1 x 21

32
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

2x
= - 3
x 2  1( x 2  1) 2

2. Find the slope of the curve y = x3 – x any point x = a. What is the slope at the point x = 1? Where
does the slope equal 11?

Solution;
y = x3 – x
dy
= 3x2 – 1
dx
At any point x = a,
dy
= 3a2 – 1
dx
The slope of the curve any point x = a is 3a2 – 1.
At the point x = 1,
dy
= 3(1)2 – 1 = 2
dx
The slope of the curve at the point x = 1 is 2.
dy
If = 11,
dx
3x2 – 1 = 11
3x2 = 12
x2 = 4
x =  2
If x = 2,
y = 23 – 2 = 6
If x = - 2,
y = (-2)3 – (-2) = - 6

At the points (2, 6) and (-2, -6), the slope equal 11.

3x 2  8
3. Calculate the slope of the curve y = at the point (2, 4).
5  2x

Solution;
3x 2  8
y =
5  2x
d d
dy (5  2 x ) (3 x 2  8)  (3 x 2  8) (5  2 x )
= dx dx
dx (5  2 x ) 2

(5  2 x )6 x  (3 x 2  8)(2)
=
(5  2 x ) 2
30 x  12 x 2  6 x 2  16
=
(5  2 x ) 2

33
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

 6 x 2  30 x  16
=
(5  2 x ) 2

at the point (2, 4),

dy  6( 2) 2  30 ( 2)  16
= = 20
dx (5  2( 2)) 2
The slope of the curve at the point (2, 4) is 20.

2x2 1
4. Find equations of the tangent line and normal line to the curve y = at the point (1, 3).
x

Solution;
2x2 1
Let y = f (x) = = 2x + x-1
x
2x 2 1
f ' (x) = 2 + (- x-2) =
x2
at the point (1, 3),
2(1) 2  1
f ' (x) = =1
12
The equation of the tangent at the point (1, 3) is 𝑦 𝑦₁ = f ' (x1) (x  x₁)
y  3 = 1 (x  1)
y3 = x1
y = x+2
1
The equation of the normal at the point (1, 3) is 𝑦 𝑦₁ = (x  x₁)
f ( x1 )
1
y3 = (x  1)
1
y3 = -x+1
y = -x+4

5. Find y' and y'' for each of the following functions.

x x 1
(a) y = (b) y = x x 2 (c) y = (d) y = (3x + 2)10
x 1 x2

Solution;
x
(a) y =
x 1
d d
( x  1) x  x ( x  1)
y' = dx dx
( x  1) 2

( x  1)(1)  x(1) 1
= = = - (x - 1)-2
( x  1) 2 ( x  1) 2

34
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

2
y'' = 2 (x - 1)-3 =
( x  1) 3

(b) y = x x 2
d dy
y' = x x2+ x2 x
dx dx
1
1 
= x ( x  2) 2 (1) + x  2 (1)
2
x
= + x2
2 x2
x  2( x  2) 3x  4
= =
2 x2 2 x2
d d
x2 (3x  4)  (3x  4) x2
1 dx dx
y'' =
2 ( x  2)2
1

1 x  2 (3)  (3 x  4) 12 ( x  2) 2 (1)
=
2 ( x  2)
3x  4
3 x2 
=
1 2 x2
2 ( x  2)
1 6 x  12  3x  4 1 3x  8
= =
2 2 x2 x2 3
4( x  2) 2

x 1
(c) y = = x-1 + x- 2
x2
1 2 x2
y' = -1x-2 - 2x-3 = - 2
- 3 =
x x x3
2 6 2x  6
y'' = 2x-3 + 6x-4 = 3
+ 4 =
x x x4

(d) y = (3x + 2)10

d
y' = 10 (3x + 2)9(3x + 2) = 10 (3x + 2)9 (3) = 30 (3x + 2)9
dx
d
y'' = 270 (3x + 2)8 (3x + 2) = 270 (3x + 2)8 (3) = 810 (3x + 2)8
dx

2x 2  3
6. If y = , prove that x2 y'' + x y' = y.
x

Solution;

35
U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

2x 2  3
y = = 2x + 3x-1
x
3 2x 2  3
y' = 2 – 3x-2 = 2- =
x2 x2
6
y'' = 6x-3 =
x3
6 2x 2  3
x2 y'' + x y' = x2 ( ) + x ( )
x3 x2
6 2x 2  3
= +
x x
2x2  3
= =y
x
 x2 y'' + x y' = y

7. If y = 2 x  1 , show that - y3 y'' + yy' = 2.

Solution;
y = 2x 1
1
1  d
y' = (2x - 1 ) 2 (2x  1)
2 dx
1 1
1   1
= (2x - 1 ) 2 (2) = (2x - 1 ) 2 =
2 2x - 1
3
1  d
y'' = - (2x - 1 ) 2 (2x  1)
2 dx
3
1  -1 1
= - (2x - 1 ) 2 (2) = 3
= -
2 (2 x  1) 2x -1
(2 x  1) 2
-1 1
- y3 y'' +yy' = - ( 2 x  1 )3 + 2x 1
(2 x  1) 2x - 1 2x - 1
= 1+1 =2
 - y3 y'' + yy' = 2

11.4 Implicit Differentiation

We have been studied the differentiation of the equation of the form y = f ( x ). However some
functions like
xy1
x 2 y + xy 2 = 7 y2 = y 2  2 xy  3  2 y
x 1

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

cannot be differentiated in a usual way. These equations define an implicit relation between the
dependent and independent variable y and x . To calculate the derivative of implicitly defined
functions, we proceed the following steps;

Suppose y as a differentiable function of x.

Step 1 Differentiable both sides of the equation with respect to x.

dy dy
Step 2 Collect the terms with on one side of the equation and solve for .
dx dx

Example 30.
dy
If x2 – xy2 – y3 = 2, find .
dx

Solution;
x2 – xy2 – y3 = 2
Differentiate with respect to x on both sides.
dy 2 dx dy
2x – [ x. + y2 . ] – 3y2 . = 0
dx dx dx
dy dy
2x – x .2y – y2 – 3y 2. = 0
dx dx
dy
(2xy + 3y2) = 2x – y2
dx
dy 2x  y 2
=
dx 2xy  3y 2

Example 31.
Find the equation of the tangent line to the curve 3x2 + 2y2 = 2xy + 23 at the point (3, 2).

Solution;
Curve: 3x2 + 2y2 = 2xy + 23
Differentiate with respect to x on both sides.
dy dy
6x + 4 y = 2[x. +y]
dx dx
dy
(4y – 2x) = 2y – 6x
dx
dy y  3x
=
dx 2y  x
The gradient of tangent to the curve at (3, 2) is
2  3(3)
m = = –7
2(2)  3

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Then equation of the tangent line is y–2 = – 7 (x – 3)

7x + y = 23

Example 32.
d2 y
If x 2  y 2  25 , find .
d x2

Solution;
x 2  y 2  25
We differentiate both side of the equation with respect to x, we get
dy
2x - 2y = 0
dx
dy x
= .
dx y
Next
dy
y x
d2 y d  x dx 1 x dy
=   = = - .
dx 2
dx  y  y2
y y2 dx
1 x x
=  .
y y2 y
1 x2
= 
y y3

Exercise 11.5
dy
1. Find .
dx
(a) xy = 5 (b) x (x + y) = y2 (c) x3 – 4 xy + y2 = 14
1 1 1
(d) 2
 2 = (e) x3 + xy + y2 = 2 (f) x  y =1
x y 4

Solution;
(a) xy = 5 ,
Differentiate with respect to x on both sides.
dy d
x +y x = 0
dx dx
dy
x +y = -y
dx
dy
x = -y
dx
dy y
= -
dx x

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

(b) x (x + y) = y2
x2 + x y = y2
Differentiate with respect to x on both sides.
dy d dy
2x + x +y x = 2y
dx dx dx
dy dy
2x + x +y = 2y
dx dx
dy dy
x - 2y = - 2x – y
dx dx
dy
(x- 2y) = - 2x - y
dx
dy  2x  y
=
dx x  2y

(c) x3 – 4 xy + y2 = 14
Differentiate with respect to x on both sides.
dy d dy
3x2 - 4 (x +y x) + 2y = 0
dx dx dx
dy dy
3x2 - 4x - 4y (1) + 2y = 0
dx dx
dy
(- 4x + 2y) = 4y - 3x2
dx
dy
= 4 y  3x
2

dx  4x  2 y

1 1 1
(d) 2
 2 =
x y 4
1
x-2 + y-2 =
4
Differentiate with respect to x on both sides.
dy
- 2x-3 + (- 2 y-3) = 0
dx
2 dy 2
- =
y3 dx x3
dy 2 y3
= (- )
dx x3 2
dy y3
= -
dx x3

(e) x3 + xy + y2 = 2
Differentiate with respect to x on both sides.

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

dy d dy
3x2 + x +y x + 2y = 0
dx dx dx
dy dy
3x2 + x + y (1) + 2y = 0
dx dx
dy dy
x + 2y = 3x2  y
dx dx
dy
( x + 2y) = 3x2  y
dx
dy  3x 2  y
=
dx x  2y

(f) x  y =1
Differentiate with respect to x on both sides.
1
1  1  1 dy
( x) 2  ( y ) 2 = 0
2 2 dx
1 1 dy
 = 0
2 x 2 y dx
1 dy 1
- = -
2 y dx 2 x

dy y y
= =
dx x x

2. Find equations of the tangent line and normal line to the crve y2 = x3 at the point (1, -1).

Solution;
y 2 = x3
Differentiate with respect to x on both sides,
dy
2y = 3x2
dx
dy 3x 2
=
dx 2y
The gradient of tagent to the curve at (1, 1) is
3(1) 2 3
m = =-
2(1) 2
The equation of the tangent line is
3
y  ( 1) = - (x  1)
2
2y + 2 = - 3x + 3
3x + 2y = 1

3. Find y' if x3 + y3 = 4xy. Find an equation of the tangent line to the curve x3 + y3 = 4xy at the point
(2, 2).

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

Solution;
x3 + y3 = 4xy
Differentiate with respect to x on both sides,
d
3x2 +3y2 y' = 4 (x y' + y x)
dx
3x2 + 3y2 y' = 4x y' + 4y (1)
y' (3y2  4x) = 4y – 3x2
4 y  3x 2
y' =
3y 2  4x
The gradient of the tangent to the curve at (2, 2) is
4( 2)  3( 2) 2 8  12
m = = =-1
2
3( 2)  4(2) 12  8
The equation of the tangent line is
y  2 = 1 (x  2)
y2 = x+2
x+y = 4

4. Find y' if x2  xy  y2 = 1. Find an equation of the tangent line to the curve x 2  xy  y2 = 1

at the point (2, 1).

Solution;
x2  xy  y2 = 1
Differentiate with respect to x on both sides,
d
2x  (x y' + y x)  2y y' = 0
dx
2x  xy'  y  2yy' = 0
y' (x  2y) =  2x + y
 2x  y 2x  y
y' = =
 x  2y x  2y
The gradient of the tangent at (2, 1) is
2( 2)  1 3
m= =
2  2(1) 4
The equation of the tangent is
3
y1 = (x  2)
4
4y  4 = 3x  6
3x + 4y = 2

5. Find y'' if x4 + y4 = 1.

Solution;
x4 + y4 = 1
Differentiate with respect to x on both sides,
4x3 + 4y3 y' = 0

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

4y3 y' = - 4x3

x3
y' = -
y3
d 3 d 3
y3 x  x3 y
y'' = - dx dx
( y3 )2
y 3 (3 x 2 )  x 3 (3 y 2 )
= -
y6
 3x 2 y 3  3x 3 y 2
=
y6

6. Show that the equation of the tangent to the curve x2 + xy + y = 0 at the point (a, b) is
x (2a + b) + y ( a + 1) + b = 0.

Solution;
x2 + x y + y = 0
at the point (a, b),
a2 + a b + b = 0 ----- (1)
Differentiate with respect to x,
d
2x + x y' + y x + y' = 0
dx
2x + x y' + y (1) + y' = 0
y' (x + 1) = - 2x - y
 2x  y
y' =
x 1
at the point (a, b),
 2a  b
y' =
a 1
The equation of the tangent at (a, b) is
 2a  b
y-b = (x - a)
a 1
ay–ab+y-b = -2ax + 2a2 – b x + a b
2ax + b x + ay + y - 2a2 - 2ab - b =0
x (2a + b) + y (a + 1) + b - b - 2a2 - 2ab - b = 0
x (2a + b) + y (a + 1) + b - 2a2 - 2ab - 2b = 0
x (2a + b) + y (a + 1) + b - 2 (a2 + a b + b) = 0
x (2a + b) + y (a + 1) + b = 0

7. Find the coordinates of the points on the curve x2 – y2 = 3xy – 39 at which the tangent is
parallel to the line x + y = 1.

Solution;
x2 – y2 = 3xy – 39 ----- (1)
Differentiate with respect to x on both sides,
2x – 2y y' = 3 (x y' + y)

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U YE MIN HTOO M.E [Y.T.U] GRADE - 11 Mathematics

2x – 2y y' = 3x y'+ 3y
– 2y y' - 3x y' = - 2x + 3y
(3x – 2y) y' = - 2x + 3y
 2x  3 y 2x  3 y
y' = =
3x  2 y 3x  2 y

x+y = 1
y = -x+1
Compare y = mx + c,
m = -1

The tangent is parallel to the line x + y = 1,

2x  3y
= -1
3x  2 y
2x – 3y = -3x – 2y
y = 5x in eq; (1)
x2 – (5x)2 = 3x (5x) – 39
x2 – 25x2 = 15x2 – 39
39x2 = 39
2
x = 1
x = 1
If x = 1, y = 5 (1) = 5
If x = -1, y = 5 (-1) = -5
The coordinates of the points on the curve at which the tangent is parallel to the line
x + y = 1 are (1, 5) and (-1, -5).

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