5128_Ch03_pp098-184

120 Chapter 3 Derivatives
EXAMPLE 4
Differentiating a Product
Find f x if f x x 2 1x 3 3.
SOLUTION
From the Product Rule with u x 2 1 and v x 3 3, we find
d
f x x d x
2 1x 3 3 x 2 13x 2 x 3 32x
3x 4 3x 2 2x 4 6x
5x 4 3x 2 6x. Now try Exercise 13.
Using the Quotient Rule
Since order is important in subtraction,
be sure to set up the numerator of the
Quotient Rule correctly:
v times the derivative of u
minus
u times the derivative of v.
You can remember the Quotient Rule
with the phrase “bottom times the derivative
of the top minus the top times
the derivative of the bottom, all over
the bottom squared.”
We could also have done Example 4 by multiplying out the original expression and
then differentiating the resulting polynomial. That alternate strategy will not work, however,
on a product like x 2 sin x.
Just as the derivative of the product of two differentiable functions is not the product of
their derivatives, the derivative of a quotient of two functions is not the quotient of their
derivatives. What happens instead is this:
RULE 6
The Quotient Rule
At a point where v 0, the quotient y uv of two differentiable functions is differentiable,
and
v
d
d
( x
u v ) d u
u d
v
dx
dx
.
v 2
Proof of Rule 6
d
d
( x
u v ) lim
h→0
vxux h uxvx h
lim
h→0
hvx hvx
To change the last fraction into an equivalent one that contains the difference quotients for
the derivatives of u and v, we subtract and add vxux in the numerator. This allows us
to continue with
( u v ) lim
d
d
x
h→0
u x
vx
h
h
u x
v
x
h
vxux h vxux vxux uxvx h
hvx hvx
vx ux h ux
ux vx h vx
h
h
lim
vx hvx
h→0
.
Taking the limits in both the numerator and denominator now gives us the Quotient
Rule.
■
EXAMPLE 5 Supporting Computations Graphically
Differentiate f (x) x x
2
2
1
. Support graphically.
1
continued