TOPIC 3 : DIFFERENTIATION

Derivative of a Function
Rules of Differentiation
Differentiation of Exponential,
Logarithmic and Trigonometric Functions

1
 Derivative of a Function

Learning Outcomes:
At the end of this topic, students should
be able to:
(a) Determine the derivative of a
function
by using the first principle.
(b) Determine the differentiability of a
function at x = a.

2
 Derivative of a Function

The process of finding the derivative of a

function is called differentiation.

We can write the differentiation as

dy d
dx
or f '
( x) or y '
or y
dx

3
 Note:

The differentiation of a function should be

with respect to the independent variable
for example, if

dy
i) y = f(x), then = f ( x)
'

dx
dy
ii) y = h(t), then = h ( t)
'

dt
4
DERIVATIVE OF A FUNCTION FROM
FIRST PRINCIPLES

Definition 1 :

For a function y = f (x), the derivative of

f (x), written f ’(x) is defined by

f(x  h)  f(x)
f'(x) = lim
h0 h

5
 Example 1 :

Use the first principle to find the

derivative of :

( a) f ( x ) = 3x

( b) f ( x ) = x2  2x

( c) f ( x) = x  5

1
( d) f ( x) =
x
6
 Solution 1(a):
( a) f ( x ) = 3x
f ( x  h) = 3( x  h)

3 ( x  h )  3x
f '
( x ) = lim
h �0 h
3x  3h  3 x
= lim
h �0 h
3h
= lim
h �0 h

=3 7
 Solution 1(b):
( b) f ( x ) = x2  2x
f ( x  h) = ( x  h)  2 ( x  h)
2

(
� ) ( ) �
2
x  h  2 x  h  �
x
�
2
 2x�
�
f ( x ) = lim
' � �
h �0 h

x  2 xh  h  2 x  2h  x  2 x
2 2 2
= lim
h �0 h

8
 Solution 1(b):

2 xh  h  2h
2
= lim
h �0 h
h ( 2x  h  2)
= lim
h �0 h

= 2x  2

9
 Solution 1(c):

( c) f ( x) = x  5

f ( x  h) = ( x  h)  5

( x  h)  5  x5
f '
( x ) = lim
h �0 h

10
 Solution 1(c):

( x  h)  5  x5 ( x  h)  5  x5
= lim �
h �0 h ( x  h)  5  x5

( ( x  h)  5 ) ( )
2 2
 x5
= lim
h �0
h ( ( x  h)  5  x5 )
= lim
( x  h  5)  ( x  5)
h �0
h ( ( x  h)  5  x  5 )
11
 Solution 1(c):
h
= lim
h �0
h ( ( x  h)  5  x5 )
1
= lim
h �0
( ( x  h)  5  x5 )
1
=
2 x5

12
 Solution 1(d):
1 1
( d) f ( x) =
x
f ( x  h) =
xh
1 1

f ( x ) = lim
' x  h x
h �0 h
x  ( x  h)
= lim
h �0 h ( x  h ) ( x )

h
= lim
h �0 h ( x  h ) ( x )

1 1
= lim  = 2
h �0 ( x  h ) ( x ) x 13
DERIVATIVE OF A FUNCTION AT x=a
Definition 2 :
The derivative of a function f(x) at any
point x = a is given by

f(x)  f(a)
f'(a) = lim
x a xa

14
 Example 2 :

f ( x)  f ( a)
Use the definition f
'
( a ) = lim
x �a xa

( a) f ( x ) = 3  2 x, a=2

( b) f ( x ) = x  2, a=3

15
 Solution 2(a):
( a) f ( x ) = 3  2 x, a=2

( 3  2x )  ( 3  2 ( 2) )
f ( 2 ) = lim
'
x �2 x2
3  2x 1
= lim
x �2 x2
4  2x 0
= lim Direct substitution gives
x �2 x  2 0

2( 2  x)
= lim
x �2 x2 16
 Solution 2(a):
2( 2  x)
= lim
x �2  ( 2  x)
= 2

17
 Solution 2(b):

( b) f ( x ) = x  2, a=3

x  2  3 2
f ( 3) = lim
'
x �3 x3

x  2 1 Direct substitution
= lim
x �3 x 3 gives
0
0
x  2 1 x  2 1
= lim �
x �3 x 3 x  2 1
18
 Solution 2(b):

= lim
( x  2) 1
x �3
( x  3) ( x  2  1)
x 3
= lim
x �3
( x  3) ( )
x  2 1

1
= lim
x �3
( x  2 1 )
1
=
2
19
 Example 3 :
2 x, x  0
Given f ( x ) =  x 2 , x  0

Is f is differentiable at x = 0 ?

Solution 3:
f ( x)  f (0) 2x  0
f '(0) = lim = lim = lim 2 = 2
x �0 x0 x �0 x x �0

f ( x)  f (0) x2  0
f '(0) = lim = lim = lim x = 0
x �0 x0 x �0 x x �0

20
Left hand limit is not equal to right
hand limit.
Therefore f '(0) does not exist.
Hence,f is not differentiable at x = 0.

21
 Example 4 :
,

Determine whether the function f ( x) = x

is differentiable at x=0.

Solution 4:
�x , x  0
f ( x) = x = �
�x , x �0

22
 Solution 4:
,

For x  0
f ( x)  f (0) x  0
f '(0) = lim = lim = lim ( 1) = 1
x �0 x0 x �0 x x �0

For x �0
f ( x)  f (0) x0
f '(0) = lim = lim = lim (1) = 1
x �0 x0 x �0 x x �0

Left hand limit is not equal to right hand

limit.
Therefore f '(0) does not exist.
Hence, f is not differentiable at x = 0.

23
 Rules of Differentiation
Learning Outcomes:
At the end of this topic, students should
be able to:
(a) Apply the rules of differentiation.
(b) Perform second and third order
differentiation

24
 Rules of Differentiation

The Constant Rule

If f (x) = k , where k is a constant , f ’(x) = 0

Example 1:
Find the derivatives of
(a) f ( x ) = 31 (b) f ( x ) = 2p

f ' ( x) = 0 f ' ( x) = 0

25
 The Power Rule
If f (x) = xn, then
f ’(x) = nxn-1 where n is positive integer.
Example 2:
Find the derivatives of
1
( a) f ( x) = x 23
( b) f ( x ) =
x
1 3
f ' ( x) =  ( x) 2

f '
( x ) = 23x 22
2
1
= 3
2x 2 26
 If f (x) = kxn, then
f ’(x) = knxn-1 where k is a constant.

Example 3:
9
Find the derivative of f ( x ) =

1 x
f ( x ) = 9x 2
� 1 2 1 �

1
f ( x) = 9�
'
 x �
�2 �
3
9  9
= x 2
= 3
2
2x 2
27
If y = [ f (x) ]n , then y ’ = n[ f (x) ]n-1 f ’(x)

Example 4:
By using the Power Rule, differentiate the
following functions :

( a) y = ( 7x  1)
2 5

( b) y = 3x  1

28
 Solution :

( a) y = ( 7x  1) ( b) y =
5
2
3x  1
1
y = ( 3x  1) 2

1 1
y = ( 3x  1) 2 ( 3)

y = 5 ( 7 x  1) ( 14 x )
4 '
' 2
2
3
= 70 x ( 7 x  1)
4 1
= ( 3 x  1) 2
2 

2
3
=
2 2x 1
29
 The Sum / Difference Rule
If f (x) = h (x) ± g(x) , then
f ’(x) = h’(x) ± g’(x)

Example 5:
Find the derivatives of the following:
( a ) f ( x ) = 3x  2 x  6 x  11
8 5

2
7x
( b) f ( x ) = 5
3
( c) f ( x ) = 2x 2
5 x
30
 Solution :
( ) ( )
a f x = 3 x 8
 2 x 5
 6 x  11
f '
( x ) = 24 x 7
 10 x  6
4

2
7x
( b) f ( x ) = 5
3
14
f ( x) = x
'

31
 Solution :

( c) f ( x ) = 2x 2
5 x
1
= 2x  5x2 2

1
5 
f ( x) = 4x  x
' 2
2
5
= 4x 
2 x

32
 The Product Rule

If y = uv, then
dy dv du
=u v = uv ' vu '
dx dx dx

Similarly, f ( x ) = u ( x ) v ( x ) , then

f '
( x) = u ( x) v ( x)  v ( x) u ( x)
' '

33
 Example 6:
By using the product rule differentiate
the following with respect to x :
� 1 �
( a ) y = ( x  1) �3x  4 �
2

� x �

( b ) y = ( x  2 ) ( 2 x  3)
4 3

34
 Solution 6 (a):
� 1 �
( a ) y = ( x  1) �3x  4 �
2

� x �

Product dy dv du
=u v
Rule dx dx dx
1
u = x 1
2
v = 3x  4
x
du
= 2x = 3x  x 4
dx
dv
= 3  4 x 5
dx
35
 Solution 6 (a):
dy
= ( x  1) ( 3  4 x )  ( 3x  x ) ( 2 x )
2 5 4

dx
2
4x 4 2x
= 3x  5  3  5  6 x  4
2 2

x x x
2 4
= 9x  3  5  3
2

x x

36
 Solution 6 (b):

( b ) y = ( x  2 ) ( 2 x  3)
4 3

Product dy
= uv�
 vu �
Rule dx

u = ( x  2) v = ( 2 x  3)
4 3

du dv
= 4 ( x  2) = 3 ( 2 x  3) ( 2 )
3 2

dx dx
= 6 ( 2 x  3)
2

37
 Solution 6 (b):
dy
= ( x  2 ) 6 ( 2 x  3)  ( 2 x  3) 4 ( x  2 ) �
� � �
4 2 3 3

dx � � � �
= ( x  2 ) ( 2 x  3) � ( ) ( )
3 2
6
� x  2  4 2 x  3 �
�
= ( x  2) ( 2 x  3) [ 6 x  12  8 x  12]
3 2

= ( x  2 ) ( 2 x  3) [ 14 x  24] Note :
3 2

= 2 ( x  2 ) ( 2 x  3) ( 7 x  12 )
3 2 6 a 4 2
b  4 a 3 3
b
= a 3b 2 ( 6a  4b )

38
 The Quotient Rule
u
If y = , then
v
du dv
v u
dy
= dx 2 dx
dx v
u ( x)
Similarly, f ( x ) = , then
v ( x)
v ( x ) u ' ( x )  u ( x ) v' ( x )
f ' ( x) =
v ( x) �
2
�
� � 39
 Example 7:

By using Quotient Rule, find dy

dx
x
( a) y =
x2

( 3x  1)
2

( b) y=
x2  1

40
 Solution 7 (a):
x du dv
v u
( a) y = Quotient dy
= dx 2 dx
x2 Rule dx v
u=x
du
=1 dy ( x  2 ) ( 1)  ( x ) ( 1)
dx =
( x  2)
2
dx

2
v = x2 =
( x  2)
2

dv
=1
dx
41
 Solution 7 (b):
( 3 x  1)
2
dy vu ' uv '
( b) y= Quotient =
x 12
Rule dx v2

u = ( 3x  1) dy ( x  1) 6 ( 3 x  1)  ( 3 x  1) ( 2x)
2 2 2

=
u ' = 2 ( 3x  1) �
3 dx ( x  1)
2 2

= 6 ( 3 x  1) 3 ( x  1)  x ( 3 x  1) �
2 ( 3 x  1) �
� 2

= �
( x  1)
2 2

v = x 1
2

2 ( 3 x  1) ( 3  x )
v ' = 2x =
(x  1)
2 2
42
 The Chain Rule

g ( x) �
If y = f �
� �and u = g ( x ) ,
then y = f ( u )

dy dy du
= �
dx du dx

43
 Example 8:

Differentiate the following with respect

to x, using the Chain Rule.

( a ) y = ( x  4)
3

1
( b) y =
4x  9

44
 Solution 8 (a) :
Chain dy dy du
= �
( a ) y = ( x  4)
3
Rule dx du dx

Let u = ( x  4 ) dy
= (3u )(1)
2

du dx
=1
dx
Substitute, u = x  4
dy
y=u 3
= 3( x  4) 2

dx
dy
= 3u 2

du
45
 Solution 8(b):
1 Chain dy dy du
( b) y = = �
4x  9 Rule dx du dx

= ( 4x  9)
1
dy �1 �
( 4)
= �2 �
dx �u �
Let u = 4 x  9
du Substitute, u = 4 x  9
=4
dx
dy 4
1 =
y=u dx ( 4 x  9 ) 2

dy 2
= u
du 46
 Example 9:
dy
Find if y = u  3u  1 and u = x  2
3 2 2

dx

Solution 9 :

y = u  3u  1
3 2
u = x 2
2

dy du
= 3u  6u
2
= 2x
du dx

47
 Solution 9:

Chain dy dy du
= �
Rule dx du dx
dy
= (3u  6u )(2 x)
2

dx
Substitute, u = x 2  2
dy
= ( 3( x 2  2) 2  6( x 2  2) ) (2 x)
dx
dy
= 6 x ( x  2)
3 2

dx
48
 Second Order Differentiation
Second order differentiation is the
differentiation of the first derivative.

Third Order Differentiation

Third order differentiation is the
differentiation of the second derivative.

49
 Example 10:
d2y d3y
Find 2
and 3
for the following:
dx dx

( a) y = 2x 3
 5x  7 x  6
2

x 5
( b) y =
x

1
( c) y =
2x  3
50
 Solution 10 (a) :

( )
a y = 2 x 3
 5 x 2
 7x  6

dy
= 6 x 2  10 x  7
dx
d2y
2
= 12 x  10
dx
3
d y
3
= 12
dx

51
 Solution 10 (b):
2 5
x 5 d y 5� 3� 
( b) y = dx 2
= � �x
2� 2�
2

x
15
x 5 =
=  5

x x 4x 2

1

= 1  5x 2 3
d y 15 � 5 � 
7

3
= �  �x 2
dx 4 � 2�
dy 5 
3 75
= x 2 = 7
dx 2 8x 2
52
 Solution 10 (c) :
1
( c) y =
2x  3
= ( 2 x  3)
1

dy
= ( 1) ( 2 x  3) ( 2 )
2

dx
= 2 ( 2 x  3)
2

2
=
( 2 x  3)
2

53
 Solution 10 (c) :
2
d y
= ( 2 ) ( 2 ) ( 2 x  3) ( 2 )
3
2
dx
8
= 8 ( 2 x  3)
3
=
( 2 x  3)
3

3
d y
= 8 ( 3) ( 2 x  3) ( 2 )
4
3
dx
48
= 48 ( 2 x  3)
4
=
( 2 x  3)
4

54
Differentiation of Exponential,
Logarithmic and Trigonometric
Functions

55
• Determine the derivative of the functions;
x f ( x) x f ( x)
i. a , a ,e ,e
ii. ln x,ln f ( x )
iii. sin x,cos x, tan x, sec x, cosec x, cot x
iv. sin u,cos u, tan u, sec u, cosec u, cot u
v. sin n x,cos n x, tan n x, sec n x, cosec n x, cot n x .

• Solve problems involving the combination of

differentiation rules.
56
 x
1. For a simplest index function, a
d x
dx
( )
a = a ln a
x

If m is a scalar,
d mx
dx
( )
a = ma ln a
mx

f ( x)
2. For the general exponential function a ,
where f(x) is a function of of x,
d
dx
a( )
f ( x)
= a ln a �f ' ( x )
f ( x)

57
 Example 1

Differentiate the following with respect to

for each of the following function:

a) y = 5 x

3 x 5
b) y = 4
c) y = 35 x
 Solution 1
a) y = 5
x
b) y = 4 3 x 5

dy d dy d
= 5 ln 5 x ln 4 ( 3 x  5)
x 3 x 5
=4
dx dx dx dx
dy dy
= 5 ln 5 = 4 ( 3) ln 4
x 3 x 5

dx dx
5 x
c) y = 3
dy d
= 3 ln 3 ( 5  x )
5 x

dx dx
dy
= 3 ( 1) ln 3 = 35 x ln 3
5 x
dx
 x
3. For a simplest index function e ,
d x
dx
( e )=e x

If m is a scalar,
d mx
dx
( )
e = me mx

f ( x)
4. For the general exponential function e ,
where f(x) is a function of of x,

dx
(
d f ( x)
e )
= f ' ( x)e f ( x)

60
 Example 2

dy
Find for each of the following function:
dx
a) y = e 4x

3 x 2 5
b) y = e
( 5 x ) 3
c) y = e
 Solution 2

a) y = e 4x
b) y = e 3 x 2 5

dy 4x d
dx
=e
dx
4x dy
dx
=e 3 x 2 5 d

dx
( 3 x 2  5)
dy 4x dy 3 x 2 5
dx
= e (4) =e ( 6x)
dx
4x
= 4e
3 x 2 5
= 6 xe
 Solution 2

( 5 x ) 3
c) y = e
dy ( 5 x ) 3 d
=e ( 5  x) 3

dx dx
dy 3
( 5 x )
=e ( 3) ( 5  x ) (1)
2
dx
3
= 3 ( 5  x ) e
2 ( 5 x )

63
 Example 3

dy
Find for each of the following function:
dx

3 2 x
a) y = xe 4x
c) y = e
x

3x  2 e x
b) y = d) y = x  x
e x
e e
 Solution 3 a)

a) y = xe 4x

dy d 4x 4x d

dx
=x
dx
( )
e e
dx
( x)
dy
= x ( 4e )  e ( 1)
4x 4x

dx
dy
= 4 xe  e4x 4x

dx
dy
= e 4 x ( 4 x  1)
dx
 Solution 3 b)
3x  2
b) y = x
e
d d x
dy
e ( 3x  2 )  ( 3x  2 ) ( e )
x

= dx dx
(e )
2
dx x

dy e x ( 3)  ( 3 x  2 ) e x
=
dx (e )
x 2

dy e ( 3  3 x  2 ) 1 3 x
x
= = x
dx (e )
x 2
e
 Solution 3 c)
3 2 x
c) y = e
x

= � e
dx x �dx
( �
�
)
dy 3 �d 2 x � 2 x �d �3 �
e � � �
�
�
�
dx �x �
�
dy 3
= ( 2 ) e
dx x
(
2 x
)
2 x �
 e �2 �
3 �
�x �
dy 2 x �6 3 �
= e �  2 �
dx �x x �
 Solution 3 d)
x
e
d) y = x  x
e e
d x x d
dy (e x
e x
) ( )dx
(e e
dx
e x  e x )
=
( )
2
dx e e
x x

=
( )
dy e x  e  x e x  e x e x  e  x x x x
e e e e e (x x ) ( )
=
dx e e
x  x 2
( e e
x )
x 2
( )
dy
=
(
e x 2e  x ) =
2

( ) (e )
dx 2 2
e x  e x x
e x
5. The natural logarithmic function is
differentiable for all x  0 .
d 1
(ln x) =
dx x

6. If the natural logarithmic function in the form

�f ( x ) �
y = ln � � where f(x) is a function of x.

d f ' ( x)
[ ln f ( x ) ] =
dx f ( x)
69
 Example 4

dy
Find for each of the following function:
dx

a) y = 5 ln x (
c) y = 7 ln 1  2 x 3
)

x2
b) y = ln( 4  3 x ) d) y = ln
1 x
 Solution 4
a) y = 5 ln x
dy 5
= (
c) y = 7 ln 1  2 x 3
)
dx x dy
=
7 d
dx 1  2 x3 dx
1  2 (
x 3
)
b) y = ln ( 4  3 x ) dy
=
(
7 6 x 2 )
dx 1  2 x3
dy 1 d
= ( 4  3x )
dx 4  3 x dx 42 x 2
=
dy 3 1  2 x3
=
dx 4  3 x
 Solution 4 d)
1
x2 �x  2 �
2 1 �x  2 �
d) y = ln = ln � � = ln � �
1 x �1  x � 2 �1  x �
1 1
y = ln ( x  2 )  ln ( 1  x )
2 2
dy 1 1 d 1 1 d
= ( x  2)  (1  x )
dx 2 ( x  2) dx 2 (1  x ) dx
dy 1 1 1 1
= [ 1]  [ 1]
dx 2 ( x  2 ) 2 (1 x)
dy 1 1 1 1
= 
dx 2 ( x  2) 2 (1  x )
dy 1  1 1  3
=    =
dx 2  x  2 1  x  2( x  2 )(1  x )
 Example 5

dy
Find for each of the following function:
dx

a) y = x ln x x
( )
c) y = e ln 2 x 3

( )
b) y = ln 4e x 2 ln 2 x
d) y = 3 x
e
 Solution 5 a) and b)
a) y = x ln x
dy �1 � u= x v = ln x
= ( x ) � � ( ln x ) ( 1)
dx �x � du dv 1
=1 =
= 1  ln x dx dx x

( )
b) y = ln 4e x2

ln ( ab ) = ln a  ln b
x2
y = ln 4  ln e = ln 4  x 2
dy ln m = p ln m
p

= 2x
dx ln e = 1
 Solution 5 c)
x
( )
c) y = e ln 2 x 3

u=e x (
v = ln 2 x 3 )
du x
dv 1 d
= 3
dx 2 x dx
2 x3 ( )
= e
dx
dv 6 x 2 3
= 3 =
dx 2 x x

dy
dx
x �3�
( )(
= e � � ln 2 x3 e  x
�x �
)
x �3
�x
( )
3 �
= e �  ln 2 x �
�
 Solution 5 d)
ln 2 x
d) y = 3 x u = ln 2 x v = e3 x
e du 1 d 1 dv
= 3e 3 x
= ( 2x) = dx
dx 2 x dx x

�1 �
dy
( e )
3x
�
�x�
� ( ln 2 x ) ( 3e 3x
)
=
dx (e )
3x 2

3x �1 �
e �  3ln 2 x �
dy
= �x � = 1  3 x ln 2 x
(e )
dx 2 3x
3x xe
Differentiation of Trigonometric
Functions

77
i) Derivatives of the functions sin x, cos x,
tan x, sec x, cosec x and cot x.

d
a) ( sin x ) = cos x
dx

d
b) ( cos x ) = -sin x
dx

d
c) ( tan x ) = sec 2 x
dx
78
i) Derivatives of the functions sin x, cos x,
tan x, sec x, cosec x and cot x.
d
d) ( sec x ) = sec x tan x
dx

d
e) ( cos ec x ) = -cos ec x cot x
dx

d
f) ( cot x ) = -cos ec x
2

dx
 Example 7
d
Prove that ( tan x ) = sec x
2

dx
Solution 7
LHS:
d d �sin x �
tan x = � � Quotient Rule
dx dx �cos x �
d d
cos x ( sin x ) - sin x ( cos x )
= dx dx
( cos x )
2
80
 cos x ( cos x ) - sin x ( - sin x )
= 2
cos x

cos x  sin x
2 2
= 2
cos x
1
= 2
cos x

= sec x
2

81
 Example 8

Differentiate with respect to x :

a) 2sin x - 3cos x

5
b) 4 tan x -
x

82
 Solution 8
d
a) ( 2sin x  3cos x ) d
b) ( 4 tan x  5 x )
1
dx
dx
= 2 cos x  3 (  sin x ) = 4sec x  ( 1) 5 x
2 2

= 2 cos x  3sin x 5
= 4sec x  2
2

x
 Example 9

Find the gradient of the following curve at a

given point :

y = 2sin x - x ,
2
x = -p

84
 Solution 9

b) y = 2sin x  x 2

dy
= 2cos x  2 x
dx
when x = p ,
dy
\ = 2 cos (p )  2(p )
dx
= 2  2p
ii) Derivatives of the function sin u, cos u,
tan u, sec u, cosec u and cot u with
u = f(x)
If u = f (x):
d d
(a) ( sin u ) = cos u u
dx dx
d d
(b) ( cos u ) = -sin u u
dx dx

d d
(c) ( tan u ) = sec u u
2

dx dx 86
ii) Derivatives of the function sin u, cos u,
tan u, sec u, cosec u and cot u with
u = f(x)

d d
(d) ( sec u ) = sec u tan u u
dx dx
d d
(e) ( cos ec u ) = -cos ec u cot u u
dx dx
d d
(f) ( cot u ) = - cos ec u u
2

dx dx
87
 Example 10

Find the derivatives of :

a) sin ( ln x )

b) cos ( sin 4x )

88
 Solution 10(a)

d
sin ( ln x ) �
�
� �
dx
d
= cos ( ln x ) ( ln x )
dx
cos ( ln x )
=
x
 Solution 10(b)

d
cos ( sin 4 x )
dx
d
=  sin ( sin 4 x ) ( sin 4 x )
dx
d
=  sin ( sin 4 x ) ( cos 4 x ) 4x
dx
= 4sin ( sin 4 x ) ( cos 4 x )
 iii) Derivatives of sinn u, cosn u, tann u,
secn u, cosecn u and cotn u with u = f(x)

If u = f (x), then
d d
a) ( sin u ) = n sin u (sin u )
n n -1

dx dx
d d
b) (cos u ) = n cos u (cos u )
n n -1

dx dx
d d
c) (tan u ) = n tan u (tan u )
n n -1

dx dx 91
 d d
d)
dx
( cot u ) = n cot u ( cot u )
n n -1

dx

d d
e)
dx
( sec u ) = n sec u (sec u )
n n -1

dx

d d
f) (cos ec u ) = n cos ec u �
n n -1
cos ec ( u ) �
� �
dx dx
92
 Example 11

Differentiate with respect to x :

a) cos 2x
4

1
b) sin ( 2 x  1)
3 2

93
 Solution 11(a)

d d
a) cos 2 x = 4 cos 2 x ( cos 2 x )
4 3

dx dx
= 4 cos 2 x ( 2sin 2 x )
3

= 8 ( cos 2 x ) ( sin 2 x )
3

94
 Solution 11(b)
d 1
b) sin 3 ( 2 x  1) 2
dx
1
d � 1
�
= 3sin 2 ( 2 x  1) 2 sin ( 2 x  1) �
�
2
dx � �
� 2 1
�� 1
�d 1
sin ( 2 x  1) 2 �
= 3� cos
� ( 2 x  1) 2
�dx ( 2 x  1) 2
� �� �
� 2 1
�� 1
�1 1
d
sin ( 2 x  1) �� cos ( 2 x  1) � ( 2 x  1) ( 2 x  1)

= 3� 2 2 2
� �� �2 dx
� 2 1
�� 1
�1 1
sin ( 2 x  1) 2 ��cos ( 2 x  1) 2 � ( 2 x  1) 2 ( 2 )

= 3�
� �� �2
� 2 1
�� 1
� 1
sin ( 2 x  1) 2 � ( ) 2 ( 2 x  1) 2

= 3� cos
� 2 x  1 �
� �� � 95
 Problem Solving involving the
combination of differentiation rules

Example 12

dy
Find for y = ( x  1)
2x

dx

96
 Solution 12
y = ( x  1)
2x

By taking ln at the both sides

ln y = ln ( x  1) 2x

ln y = 2 x ln ( x  1)
Differentiate both side toward x
1 dy d d
= 2x � ln ( x  1) �
� � ln ( x  1) ( 2 x )
y dx dx dx
�1 d �
= ( 2x) � ( x  1) � �
ln ( x  1) �
� �( 2)
�x  1 dx �
 Solution 12
1 dy �1 �
= ( 2x) � ( 1) � �
ln ( x  1) �
� ( 2)
�
y dx �x 1 �

1 dy 2x
=  2 ln ( x  1)
y dx x  1
dy �2 x �
\ = y �  2 ln ( x  1) �
dx �x  1 �
dy 2 x �2 x �
\ = ( x  1) �  2ln ( x  1) �
dx �x  1 �
98
 Example 13

Find f '( x) if f ( x) = ln ( sin x ) .

2

99
 Solution 13
d 1 d
dx
ln ( sin x ) =
2
2
sin x dx
( sin x )
2

1 d
= 2 ( 2sin x ) ( sin x )
sin x dx
1
= 2 ( 2sin x ) cos x
sin x
2 cos x
=
sin x
= 2 cot x 100
 Example 14

dy
Find if y = cos ( e )  e cos x .
x x
dx

101
 Solution 14
y = cos ( e )  e cos x
x x

dy d x x d d x
=  sin ( e )
x
( e ) e ( cos x )  cos x ( e )
dx dx dx dx

=  sin ( e ) e  e (  sin x )  ( cos x ) ( e

x x x x
)
= e sin ( e )  e sin x  e cos x
x x x x

102
 Example 15
dy
Find if y = sec ( e .)
x

dx

Solution 15
y = sec ( e x
)
dy d x
= sec ( e ) tan ( e )
x x
( e )
dx dx

= e x sec ( e x ) tan ( e x )
103