DN1.5: CHAIN RULE
The ‘chain rule’ is used to differentiate a function which is the composition of two simpler functions
If y = g[u] where u = h(x)
dy
dy
du
=
×
dx
du
dx
Then
Examples
Differentiate y = (2x - 1)4
1)
Let u = 2x – 1, then y = u4
du
dy
= 2 and
= 4u3
dx
du
dy
dy
du
=
×
dx
du
dx
= 4u3 . 2
= 8u3
[since u = 2x – 1]
= 8(2x – 1)3
2)
Find the derivative of y =
y = (5t2 + 2t + 1)
1
3
−1
3
5t + 2t + 1
2
[change to index form for easier differentiation]
Let u = 5t2 + 2t + 1, then y =
1
=u
3
u
−1
3
−4
du
dy
−1 3
=
u
= 10t + 2 and
dt
du
3
dy
dy
du
=
×
dt
du
dt
=
=
=
DN1.5 –Chain Rule
−1
3
−1
3
u
−4
3
. (10t + 2)
(5t2 + 2t + 1)
−4
3
. (10t + 2)
−4
− (10t + 2) 2
(5t + 2t + 1) 3
3
Page 1 of 3
[since u = 5t2 + 2t + 1]
[after simplifying]
June 2012
Differentiate y = sin5x
3)
y = sin5x
Let y = sin(u) where u = 5x
dy
du
= cos(u) and
=5
du
dx
Then
dy
dy
du
=
×
dx
du
dx
= cos(u). 5
= 5cos(5x)
4).
If f(x) = cos3x find f'(x)
y = cos3x = [cos(x)]3
Let y = u3 where u = cos(x)
dy
du
= 3u2 and
= -sin(x)
du
dx
Then f'(x) =
dy
dy
du
=
×
dx
du
dx
= 3u2.(-sinx)
= 3cos2x.(-sinx)
= -3sinxcos2x
5)
Differentiate (loge4x)3
Let y = u3
where u = logev and v = 4x
[The chain rule can be extended
to three or more functions!!]
dy
dv
du 1
= 3u2 ,
=
and
=4
du
dx
dv v
dy
dy du dv
=
.
.
dx
du dv dx
1
= 3u2. .4
v
1
= 3(logev)2.
.4
4x
1
= 3(loge4x)2.
.4
4x
3
=
(loge4x)2
x
DN1.5 –Chain Rule
Page 2 of 3
June 2012
Exercise
Find the derivatives of the following functions
1)
y = tan3x
3)
y = sin 
5)
f(x) = e
x

2
2) f(x) = loge 
π

− 2x 
4

4) y = cos2x
sin x
6) y =
1− cos 5 x
Answers
1) 3sec23x
2)
1
x
3) -2cos (
π
− 2 x)
4) -2 sinx cosx
4
5) esinx cosx
DN1.5 –Chain Rule
6)
5 sin 5 x
2 1 − cos 5 x
Page 3 of 3
June 2012