11.4 The Chain Rule
Composite Functions
Definition: A function m is a composite of functions f and g if
m(x) = f [g(x)]
The domain of m is the set of all numbers x such that x is in the
domain of g and g(x) is in the domain of f.
Example 1
Given f(u) = 2u and g(x) = ex
Find f [g(x)] and g [f(u)]
f [g(x)] = f (ex) substitute g(x) for ex
= 2(ex) plug ex into function f(u)
= 2ex
g [f(u)] = g (2u) substitute f(u) for 2u
= e2u plug 2u into function g(x)
Example 2
Write each function as a composite of two simpler
functions
A) Y = 50 e-2x
g(x) = -2x
and
B)
f(u) = 50 eu
y  3 1  x3
g ( x)  1  x 3 and f (u )  3 u
C) Write y as a composite of two simpler functions
y = (2x + 1)100
g(x) = 2x + 1
and f(u) = u 100
How do we find the derivative of a composite
function such as the one above?
What is the derivative of (2x + 1)100
Let’s investigate on the next slide!
Find the derivative of (2x + 1)2, (2x + 1)3, (2x + 1)4 and (2x + 1)100
Use the product rule:
F(x) = (2x + 1)2 = (2x + 1) (2x + 1)
F’(x) = (2)(2x+1) + (2)(2x+1) = 4(2x+1)
F(x) = (2x + 1)3 = (2x + 1) (2x + 1)2 = (2x+1)(4x2 + 4x +1)
F’(x) = (2)(4x2 + 4x +1) + (8x + 4)(2x + 1)
=
2 (2x+1)2
+ 4(2x+1)(2x+1)
=
2 (2x+1)2
+ 4(2x+1)(2x+1)
=
2 (2x+1)2
+ 4(2x+1)2 = 6 (2x+1)2
Try (2x + 1)4 you should find out that f’(x) = 8 (2x+1)3
Look at the relationship between f(x) and f’(x), do you agree that
the derivative of (2x + 1)100 is 200 (2x+1)99 ?
f(x)
f’(x)
(2x + 1)2
4(2x+1) = 2(2x+1) · 2
(2x + 1)3
6(2x+1)2 = 3(2x+1)2 · 2
(2x + 1)4
8(2x+1)3 = 4(2x+1)3 · 2
(2x + 1)100
200(2x+1)99 = 100(2x+1)99 · 2
Any composite function [g(x)] n n·[g(x)]n-1 · g’(x)
Generalized Power Rule
What we have seen is an example of the generalized power
rule: If u is a function of x, then
d n
n 1 du
u  nu
dx
dx
Chain Rule
We have used the generalized power rule to find derivatives of
composite functions of the form f (g(x)) where f (u) = un is a power
function. But what if f is not a power function? It is a more
general rule, the chain rule, that enables us to compute the
derivatives of many composite functions of the form f(g(x)).
Chain Rule: If y = f (u) and u = g(x) define the composite
function y = m(x) = f [g(x)], then
dy dy du
dy
du


, provided
and
exist .
dx du dx
du
dx
Or m’(x) = f’[g(x)] g’(x) provided that f’[g(x) and g’(x) exist
Example 3
Find the indicated derivatives
A) h’(x) if h(x) = (5x + 2)3
h’(x) = 3 (5x + 2)2 (5x+2)’
= 3 (5x + 2)2 (5) =15 (5x + 2)2
B) y’ if y = (x4 – 5)5
y’= 5(x4 – 5)4 (x4 – 5)’
= 5(x4 – 5)4 (4x3) = 20x3 (x4 – 5)4
Example 3
Find the indicated derivatives

d
1
d
2
C)

t
4
2
2
dt t  4
dt


D) dg if g ( w)  4  w
dw

2


3
 2 t  4 (2t ) 
2
 (4  w)
dg 1
 4  w
dw 2
1
2
t
 4t
2
4
1
2
 1

1
24  w
1
2

3
Example 4
Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x)
A) y = u-5 and u = 2x3 + 4
dy
5
6
 5u  6
du
u
du
 6x 2
dx
2
2
dy dy du

30
x

5

30
x


 6  (6 x 2 )  6  3
6
u
dx du dx
(
2
x

4
)
u
The problem above can be solved by
taking the derivative of y = (2x3 + 4) -5
Example 4 (continue)
Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x)
B) y = eu and u = 3x4 + 6
dy
 eu
du
du
 12x 3
dx
dy dy du
3 u
u
3



12
x
e  12 x 3e3 x
 e  (12 x )
dx du dx
This problem can be solved by taking the derivative of
4
6
ye
3 x 4 6
Example 5
Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x)
C) y = ln u and u = x2 + 9x + 4
dy 1

du u
du
 2x  9
dx
dy dy du 1


  ( 2 x  9)  2 2 x  9
dx du dx u
x  9x  4
This problem can be solved by taking the derivative of y = ln (x2 + 9x + 4)
The chain rule can be extended to compositions
of three or more functions.
Given: Y = f(w), w = g(u), and u = h(x)
Then
dy dy dw du



dx dw du dx
Example 6
y = h(x) = [ ln(1 + ex) ]3 find dy/dx
Let y = w 3, w = ln u, and u = 1 + ex
then dy dy dw du
dx



dw du dx
dy
2 1
x
2 1 x
 3w   e  3(ln u ) e
u
dx
u
1
x
 3[ln( 1  e )] 

e
1 ex
x
2
3e x [ln( 1  e x )]2

1 ex
Example 6 (continue)
y = h(x) = [ ln(1 + ex) ]3 find dy/dx
Another way
dy
1
x 2
x
 3 ln( 1  e ) 

(
e
)
x
dx
1 e

Derivative
of the
outside
function

The
derivative of
the function
inside the [ ]
3e x [ln( 1  e x )]2

1 ex
Derivative of
the function
inside ( )
Examples for the Power Rule
Chain rule terms are marked:
y  x5 , y'  ?
y  (2 x) 5 , y '  ?
y  (2 x 3 )5 , y '  ?
y  (2 x  1) 5 , y '  ?
y  (e x ) 5 , y '  ?
y  (ln x) 5 , y '  ?
Compare with the next slide
Examples for the Power Rule
Chain rule terms are marked:
y  x5 , y'  5x 4
y  (2 x) 5 , y '  5(2 x) 4 (2)  160 x 4
y  (2 x 3 ) 5 , y '  5(2 x 3 ) 4 (6 x 2 )  480 x14
y  (2 x  1) 5 , y '  5(2 x  1) 4 (2)  10(2 x  1) 4
y  (e x ) 5 , y '  5(e x ) 4 (e x )  5e 5 x
4
5
(ln
x
)
y  (ln x) 5 , y '  5(ln x) 4 (1 / x) 
x
Examples for
Exponential Derivatives
d u
u du
e e 
dx
dx
y  e , y'  ?
3x
ye
3 x 1
ye
4 x 2 3 x  5
, y'  ?
, y'  ?
y  e , y'  ?
ln x
Compare with the next slide
Examples for
Exponential Derivatives
d u
u du
e e 
dx
dx
y  e3 x , y '  e 3 x (3)  3e3 x
y  e3 x 1 , y '  e3 x 1 (3)  3e 3 x 1
ye
4 x 2 3 x  5
, y'  e
y  e ln x , y '  e ln x
4 x 2 3 x  5
(8 x  3)
1
e ln x x
( )
 1
x
x
x
Property of ln
Examples for
Logarithmic Derivatives
d
1 du
ln u  
dx
u dx
y  ln( 4 x), y '  ?
y  ln( 4 x  1), y '  ?
y  ln( x ),
2
y'  ?
y  ln( x  2 x  4),
2
Compare with the next slide
y'  ?
Examples for
Logarithmic Derivatives
d
1 du
ln u  
dx
u dx
1
1
y  ln( 4 x), y ' 
4 
4x
x
1
4
y  ln( 4 x  1), y ' 
4 
4x 1
4x 1
1
2
2
y  ln( x ), y '  2  (2 x) 
x
x
1
2x  2
2
y  ln( x  2 x  4), y '  2
 (2 x  2)  2
x  2x  4
x  2x  4