Section 3.2 The Product and Quotient rules
The derivative of a product of functions is not the product of the derivatives.
Examples:
a)
f (x)  2x  3
g(x)  3
f (x)g(x)  6x  9
f (x)  x  1
b)
f '(x)  2
2
f '( x)g '( x)  0
( fg )' ( x )  6
g(x)  x
f (x)g(x)  x
g '( x)  0
 x
f '( x)  1
g '( x)  1
( fg )' ( x )  2 x  1
To differentiate a product, use the product rule: (fg)' = f 'g + fg'
or (first)'(second) + (first)(second)'.
Examples: Use the shift rule also when needed.
h ( x )  ( 2 x  3 )( 4 x  5 )
1.
Check
g ( 2 x  3 )( 4 x  5 ) we have
it : Multiplyin
h(x)  8x
2
 2 x  15
2
2.
h( x)  x ( x  2)
3.
h ( x )  xe
4.
h(x)  5x
x
3
h ' ( x )  16 x  2
h '( x)  2 x( x  2)
h '( x)  e
x 1
h ' ( x )  2 ( 4 x  5 )  ( 2 x  3 ) 4  16 x  2
x
 xe
h '( x)  5
3
2
 x 3( x  2 )
2
 2 x( x  2)
3
2
 3x ( x  2)
2
x
x  1  5 x[
1
( x  1)
1 2
] 5
5x
x 1 
2
2
x 1
The derivative of a quotient is not the quotient of the derivatives.
d
Example:
f (x) 
1
 x
1
f '( x)   x
2
1

x
x
2
but
dx
d
(1 )

(x)
0
 0
.
1
dx
To differentiate a quotient use the quotient rule:
or

 f 
f ' g  fg '

 
2
g
 g 

T ' B  TB '
T 

 
2
 B 
B
Examples: Also use the shift rule where needed.
1.
h(x) 
x
2
 5x
4x  3
h'( x) 
( 2 x  5 )( 4 x  3 )  ( x
(4 x  3)
2
 5 x )4
2
Never sqauare out the denominator, leave it as it is. Only simplify the numerator.
2.
h(x) 
( x  1)
e
x
2
h'( x) 
2 ( x  1 )( e
 x
x
2
 x )  ( x  1) ( e
(e
x
 x)
x
 1)
2
3. This example shows some quotients can be written as products and then differentiated
using the product rule.
h(x) 
e
x
Re write
h(x)  x
x
Using the quotient rule
h'( x) 
xe
x
x
 e
2
x
1
e
x
Use the product
rule ,
h'( x)   x
2
e
x
 x
1
e
x