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3.2The Product and
Quotient Rules
DIFFERENTIATION RULES
In this section, we will learn about:
Formulas that enable us to differentiate new functions
formed from old functions by multiplication or division.
By analogy with the Sum and Difference
Rules, one might be tempted to guess—as
Leibniz did three centuries ago—that the
derivative of a product is the product of the
derivatives.
However, we can see that this guess is wrong by looking at a particular example.
THE PRODUCT RULE
Let f(x) = x and g(x) = x2.
Then, the Power Rule gives f’(x) = 1 and g’(x) = 2x.
However, (fg)(x) = x3.
So, (fg)’(x) = 3x2.
Thus, (fg)’ ≠ f’ g’.
THE PRODUCT RULE
If f and g are both differentiable, then:
In words, the Product Rule says: The derivative of a product of two functions is
the first function times the derivative of the second function plus the second function times the derivative of the first function.
( ) ( ) ( ) ( ) ( ) ( )d d df x g x f x g x g x f x
dx dx dx
THE PRODUCT RULE
a. If f(x) = xex, find f ’(x).
b. Find the nth derivative, f (n)(x)
THE PRODUCT RULE Example 1
By the Product Rule, we have:
THE PRODUCT RULE Example 1 a
'( ) ( )
( ) ( )
1
( 1)
x
x x
x x
x
df x xe
dxd d
x e e xdx dx
xe e
e x
Using the Product Rule again, we get:
''( ) ( 1)
( 1) ( ) ( 1)
( 1) 1
( 2)
x
x x
x x
x
df x x e
dxd d
x e e xdx dx
x e e
e x
THE PRODUCT RULE Example 1 b
Further applications of the Product Rule give:
4
'''( ) ( 3)
( ) ( 4)
x
x
f x x e
f x x e
THE PRODUCT RULE Example 1 b
In fact, each successive differentiation adds another term ex.
So:( ) ( )n xf x x n e
THE PRODUCT RULE Example 1 b
Differentiate the functionTHE PRODUCT RULE Example 2
( ) ( )f t t a bt
Using the Product Rule, we have:
THE PRODUCT RULE E. g. 2—Solution 1
1 21
2
'( ) ( ) ( )
( )
( ) 2 ( )
2 2 2( 3 )
2
d df t t a bt a bt t
dt dt
t b a bt t
a bt bt a btb t
t t ta bt
t
If we first use the laws of exponents to
rewrite f(t), then we can proceed directly
without using the Product Rule.
This is equivalent to the answer in Solution 1.
1 2 3 2
1 2 1 2312 2
( )
'( )
f t a t bt t at bt
f t at bt
THE PRODUCT RULE E. g. 2—Solution 2
( ) ( )f t t a bt
If , where
g(4) = 2 and g’(4) = 3, find f’(4).
THE PRODUCT RULE Example 3
( ) ( )f x xg x
Applying the Product Rule, we get:
So,
THE PRODUCT RULE Example 3
1 212
'( ) ( ) ( ) ( )
( )'( ) ( ) '( )
2
d d df x xg x x g x g x x
dx dx dxg x
xg x g x x xg xx
(4) 2'(4) 4 '(4) 2 3 6.5
2 22 4
gf g
If f and g are differentiable, then:
In words, the Quotient Rule says: The derivative of a quotient is the denominator times
the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
2
( ) ( ) ( ) ( )( )
( ) ( )
d dg x f x f x g xd f x dx dx
dx g x g x
THE QUOTIENT RULE
2'
3
2,
6
x xIf y find y
x
THE QUOTIENT RULE Example 4
Then,
THE QUOTIENT RULE Example 4
3 2 2 3
23
3 2 2
23
4 3 4 3 2
23
4 3 2
23
6 2 2 6'
6
6 2 1 2 3
6
2 12 6 3 3 6
6
2 6 12 6
6
d dx x x x x x
dx dxyx
x x x x x
x
x x x x x x
x
x x x x
x
Find an equation of the tangent
line to the curve y = ex / (1 + x2)
at the point .
THE QUOTIENT RULE Example 5
11,2e
According to the Quotient Rule,
we have:
THE QUOTIENT RULE Example 5
2 2
22
22
2 22 2
1 1
1
1 2 1
1 1
x x
x x x
d dx e e x
dy dx dxdx x
x e e x e x
x x
So, the slope of the tangent line at
(1, ½e) is:
This means that the tangent line at (1, ½e) is horizontal and its equation is y = ½e.
1
0
x
dy
dx
THE QUOTIENT RULE Example 5
In the figure, notice that the function
is increasing and crosses its tangent line
at (1, ½e).
THE QUOTIENT RULE Example 5
Though it is possible to
differentiate the function
using the Quotient Rule, it is much easier
to perform the division first and write
the function as
before differentiating.
23 2( )
x xF x
x
1 2( ) 3 2F x x x
NOTE
Here’s a summary of the differentiation
formulas we have learned so far.
1
'
2
0
(ln ) ' ' ' ' '
' '' ' ' ' ' '
n n x x
x x
d d dc x nx e e
dx dx dx
da a e cf cf f g f g
dx
f gf fgf g f g fg fg gf
g g
DIFFERENTIATION FORMULAS
