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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

Section 2.4 • Differentiating the Product and the Quotient of Two Functions; Higher-Order Derivatives 205

At this point, we have found the derivative, but it is customary to simplify the
answer. Then
2
2 x
x
′
y = (1 + x + 2x)e = (x + 1) e
↑ ↑
x
Factor out e . Factor.
R
NOW WORK Problem 9 and AP Practice Problem 4.
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Do not use the Product Rule unnecessarily! When one of the factors is a constant,
use the Constant Multiple Rule. For example, it is easier to work
d 2 d 2
[5(x + 1)] = 5 (x + 1) = 5 · 2x = 10x
dx dx
than it is to work

d 2 d 2 d 2 2
[5(x + 1)] = 5 (x + 1) + 5 (x + 1) = 5 · 2x + 0 · (x + 1) = 10x
dx dx dx
2
Also, it is easier to simplify f (x) = x (4x − 3) before finding the derivative. That is, it
is easier to work
d 2 d 3 2 2
[x (4x − 3)] = (4x − 3x ) = 12x − 6x
dx dx
than it is to use the Product Rule

d 2 2 d d 2 2
[x (4x − 3)] = x (4x − 3) + x (4x − 3) = (x )(4) + (2x)(4x − 3)
dx dx dx
2
2
2
= 4x + 8x − 6x = 12x − 6x
EXAMPLE 2 Differentiating a Product in Two Ways
CALC CLIP
3
2
Find the derivative of F(v) = (5v − v + 1)(v − 1) in two ways:
(a) By using the Product Rule.
(b) By multiplying the factors of the function before finding its derivative.
Solution
3
2
(a) F is the product of the two functions f (v) = 5v − v + 1 and g(v) = v − 1. Using
the Product Rule, we get
d 3 d 2 3

2
′
F (v) = (5v − v + 1) (v − 1) + (5v − v + 1) (v − 1)
dv dv
2 2 3
= (5v − v + 1)(3v ) + (10v − 1)(v − 1)
4
2
4
3
3
= 15v − 3v + 3v + 10v − 10v − v + 1
2
3
4
= 25v − 4v + 3v − 10v + 1
(b) Here we multiply the factors of F before differentiating.
2
3
4
5
3
2
F(v) = (5v − v + 1)(v − 1) = 5v − v + v − 5v + v − 1
Then
4
3
2
′
F (v) = 25v − 4v + 3v − 10v + 1
Notice that the derivative is the same whether you differentiate and then simplify,
or whether you multiply the factors and then differentiate. Use the approach that you
find easier.
NOW WORK Problem 13.
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