Business Mathematics
Lecture 7
Chain Rule
Prepared By: Dr C.S Marange & Mr R.T Chiruka
University of Fort Hare
Department of Statistics
Basic Rules of Differentiation – Rule 8
1. Chain Rule
The Chain Rule: Formula
dy dy du


dx du dx
Illustrative Example
Deriving Composite Functions using the Chain Rule
f ( x)   x  1
2
 Consider the function
2
 To compute f ’(x), we can first expand f (x)
f ( x)   x  1   x 2  1 x 2  1
2
2
 x4  2 x2  1
and then differentiate the resulting polynomial
f ( x)  4 x3  4 x
 But how should we derive a function like…
g ( x)   x  1
2
100
Deriving Composite Functions using the Chain Rule
To find the derivative of the composite function g (x):
 We let u = x2 + 1 and y = u100.
 Then we find the derivatives of each of these functions
du
 2x
dx
and
dy
 100u 99
du
 The ratios of these derivatives suggest that
dy dy du


 100u 99 .2 x
dx du dx
 Substituting x2 + 1 for u we get
99
dy
2
g ( x) 
 100  x  1 .2 x  200 x( x 2  1)99
dx
Practice Exercise
Find the derivative (solutions to follow)
1) f ( x)  (3x  5x 2 )7
2) f ( x)  3 ( x 2  1)2
3) f (t ) 
7
(2t  3) 2
Solutions
1) f ( x)  (3x  5x 2 )7
du
u  3x  5 x 
 3  10 x
dx
dy
y  u7 
 7u 6
du
2
dy dy du

dx du dx
dy
 7u 6 (3  10 x)
dx
dy
 7(3x  5 x 2 )6 (3  10 x)
dx
Solutions
2) f ( x)  3 ( x 2  1)2
u  x2  1 
y
2
 u3

du
 2x
dx
dy 2
 u
du 3
dy 2
 u
dx 3

1
3
1
3
(2 x)
dy 2 2
 ( x  1)
dx 3
dy

dx


4x
1
3( x 2  1) 3
1
3
(2 x)
Solutions
3) f (t ) 
7
(2t  3) 2
 7(2t  3) 2
du
u  2t  3 
2
dt
dy
y  7u 2 
 14u 3
du
dy dy du

dt du dt
dy
 14u 3 2
dt
dy
28
3
 14(2t  3) (2) 
dt
(2t  3)3
Class Work
Use a substitution to differentiate each
function (Exercise 2E)
1) f ( x)  (3x  1)2
2) f ( x)  (3x 2  4)
3) f (t ) 
1
(t 4  4)3
Thank you