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) y 
3
1 x
f(u) = 50 eu
3
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
dx
u  nu
n
n 1
du
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


,
dx
du dx
provided
dy
du
and
du
exist .
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
C)
D)
d
1

dt t  4
dg
2

2

d
dt
t
2
4

2

 2 t  4
2

3
( 2t ) 
 4t
t
2
4
1
if
g (w) 
4w
 (4  w) 2
dw
1
dg
dw

1
2
4  w   1
2

1
1
2 4  w 
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
  5u
6

du
du
5
u
 6x
6
2
dx
dy
dx

dy
du

du
dx

5
u
6
2
 (6 x ) 
 30 x
u
The problem above can be solved by
taking the derivative of y = (2x3 + 4) -5
6
2

 30 x
2
(2 x  4)
3
6
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
e
u
du
du
 12 x
3
dx
dy
dx

dy
du

du
dx
 e  (12 x )  12 x e
u
3
3
u
 12 x e
3
4
3x 6
This problem can be solved by taking the derivative of y  e
4
3x 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

du
du
1
u
 2x  9
dx
dy
dx

dy
du

du
dx

1
u
 (2 x  9) 
2x  9
x  9x  4
2
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
dx
 3w 
2
1
e
x
 3 (ln u )
2
1
x
u
u
 3[ln( 1  e )] 
x
x
1 e
x
1
2
3 e [ln( 1  e )]
x

e
1 e
2
x
e
x
Example 6 (continue)
y = h(x) = [ ln(1 + ex) ]3 find dy/dx
Another way
dy


dx
Derivative
of the
outside
function
1 e
 (e )
x
x
The
derivative of
the function
inside the [ ]
3 e [ln( 1  e )]
x

1
2
 3 ln( 1  e ) 
x
x
1 e
x
2
Derivative of
the function
inside ( )
Examples for the Power Rule
Chain rule terms are marked:
y  x , y' ?
5
y  (2 x) , y '  ?
5
y  (2 x ) , y '  ?
3
5
y  ( 2 x  1) , y '  ?
5
y  (e ) , y '  ?
x
5
y  (ln x ) , y '  ?
5
Compare with the next slide
Examples for the Power Rule
Chain rule terms are marked:
y  x , y' 5x
5
4
y  ( 2 x ) , y '  5 ( 2 x ) ( 2 )  160 x
5
4
4
y  ( 2 x ) , y '  5 ( 2 x ) ( 6 x )  480 x
3
5
3
4
2
14
y  ( 2 x  1) , y '  5 ( 2 x  1) ( 2 )  10 ( 2 x  1)
5
4
y  (e ) , y '  5(e ) (e )  5e
x
5
x
4
x
5x
y  (ln x ) , y '  5 (ln x ) (1 / x ) 
5
4
5 (ln x )
x
4
4
Examples for
Exponential Derivatives
d
e
u
dx
y  e , y' ?
3x
ye
3 x 1
ye
4 x 3 x5
ye
ln x
2
, y' ?
, y' ?
, y' ?
Compare with the next slide
 e 
u
du
dx
Examples for
Exponential Derivatives
d
e
 e 
u
u
dx
dx
y  e , y' e
3x
ye
3 x 1
ye
4 x 3 x5
ye
2
ln x
du
3x
, y' e
(3)  3e
3 x 1
(3)  3e
, y' e
, y' e
ln x
3x
2
4 x 3 x5
1
( )
x
e
3 x 1
(8 x  3 )
ln x

x
x
1
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  ln( x ),
2
y' ?
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
y  ln( 4 x ),
1
y'
4 
4x
y  ln( 4 x  1),
y  ln( x ),
2
y'
y'
1
x
y  ln( x  2 x  4 ),
2
2
1
x
1
4x 1
4 
 (2 x) 
4
4x 1
2
x
y'
1
x  2x  4
2
 (2 x  2) 
2x  2
x  2x  4
2