Chapter 4
Additional
Derivative Topics
Section 3
Derivatives of Products
and Quotients
Objectives for Section 4.3
Derivatives of Products and Quotients
The student will be able to
calculate:
■ the derivative of a product of
two functions, and
■ the derivative of a quotient
of two functions.
Barnett/Ziegler/Byleen Business Calculus 12e
2
Review
Exponential & Log Derivatives
Find the derivative of each function:
𝑦 = 𝑒𝑥
𝑦′ = 𝑒 𝑥
𝑦 = ln 𝑥
1
𝑦′ =
𝑥
𝑦 = 𝑏𝑥
𝑦′ = 𝑏 𝑥 ln 𝑏
𝑦 = log 𝑏 𝑥
1
𝑦′ =
𝑥 ln 𝑏
Barnett/Ziegler/Byleen Business Calculus 12e
3
Example 1
Derivative of a Product
Sometimes the derivative of a product can be found
by first using the distributive property.
Find the derivative of y = 5x2(x3 + 2).
𝑦 = 5𝑥 5 + 10𝑥 2
𝑦 ′ = 25𝑥 4 + 20𝑥
But what about a product like this one?
𝑦 = 5𝑥 8 𝑒 𝑥
You MUST use the Product Rule…
Barnett/Ziegler/Byleen Business Calculus 12e
4
Derivatives of Products
Theorem 1 (Product Rule)
If f (x) = L(x)  R(x), and if L(x) and R(x) exist, then
f (x) = L(x)  R(x) + R(x)  L(x)
Mnemonic: Left D Right + Right D Left
𝐿𝑅′ + 𝑅𝐿′
Barnett/Ziegler/Byleen Business Calculus 12e
5
Example 2
 𝑓 𝑥 = 5𝑥 8 𝑒 𝑥
Find 𝑓′(𝑥)
𝐿 𝑥 = 5𝑥 8
𝑅 𝑥 = 𝑒𝑥
𝐿′ 𝑥 = 40𝑥 7
𝑅′ 𝑥 = 𝑒 𝑥
𝑓 ′ (𝑥) = 𝐿 𝑥 ∙ 𝑅′ 𝑥 + 𝑅(𝑥) ∙ 𝐿′(𝑥)
𝑓 ′ (𝑥) = 5𝑥 8 ∙ 𝑒 𝑥 + 𝑒 𝑥 ∙ 40𝑥 7
𝑓 ′ (𝑥) = 5𝑥 7 𝑒 𝑥 (𝑥 + 8)
Barnett/Ziegler/Byleen Business Calculus 12e
6
Example 3
 𝑦 = 𝑥 + 2 3𝑥
Find y′
𝐿 𝑥 =𝑥+2
𝑅 𝑥 = 3𝑥
𝐿′ 𝑥 = 1
𝑅′ 𝑥 = 3𝑥 ln 3
𝑦 ′ = 𝐿 𝑥 ∙ 𝑅′ 𝑥 + 𝑅(𝑥) ∙ 𝐿′(𝑥)
𝑦 ′ = 𝑥 + 2 (3𝑥 ln 3) + 3𝑥 ∙ 1
𝑦 ′ = 𝑥 + 2 (3𝑥 ln 3) + 3𝑥
𝑦 ′ = 3𝑥 [ 𝑥 + 2 ln 3 + 1]
Barnett/Ziegler/Byleen Business Calculus 12e
7
Derivatives of Quotients
 Some derivatives of quotients can be found by rewriting
the function:
4𝑥 3 − 8𝑥 2
𝑓 ′ 𝑥 = 4𝑥 − 4
𝑓 𝑥 =
2𝑥
= 2𝑥 2 − 4𝑥
 You can’t do that with functions like this:
𝑡 3 − 3𝑡
𝑦= 2
𝑡 −4
Barnett/Ziegler/Byleen Business Calculus 12e
8
Derivatives of Quotients
Theorem 2 (Quotient Rule)
𝐻(𝑥)
If f (x) =
, and if H(x) and L(x) exist, then
𝐿(𝑥)
′ 𝑥 − 𝐻(𝑥) ∙ 𝐿′(𝑥)
𝐿
𝑥
∙
𝐻
𝑓′ 𝑥 =
𝐿(𝑥)2
Mnemonic:
𝐿𝑜𝑤 𝐷 𝐻𝑖𝑔ℎ−𝐻𝑖𝑔ℎ 𝐷 𝐿𝑜𝑤
𝐿𝑜𝑤 2
𝐿𝐻 ′ − 𝐻𝐿′
𝐿2
Barnett/Ziegler/Byleen Business Calculus 12e
9
Example 4
𝑡 3 − 3𝑡
Find the derivative of: 𝑦 = 2
𝑡 −4
𝐻 𝑡 = 𝑡 3 − 3𝑡
𝐿 𝑡 = 𝑡2 − 4
𝐻 ′ 𝑡 = 3𝑡 2 − 3
𝐿′ 𝑡 = 2𝑡
′ 𝑡 − 𝐻(𝑡) ∙ 𝐿′(𝑡)
𝐿
𝑡
∙
𝐻
𝑦′ =
𝐿(𝑡)2
𝑦′
𝑡 2 − 4 3𝑡 2 − 3 − 𝑡 3 − 3𝑡 2𝑡
=
𝑡2 − 4 2
4 − 3𝑡 2 − 12𝑡 2 + 12 − 2𝑡 4 − 6𝑡 2
3𝑡
𝑦′ =
𝑡2 − 4 2
Barnett/Ziegler/Byleen Business Calculus 12e
4 − 9𝑡 2 + 12
𝑡
𝑦′ =
𝑡2 − 4 2
10
Example 5
Find the derivative of 𝑦 =
𝐻 𝑥 = ln 𝑥
1
′
𝐻 𝑥 =
𝑥
𝐿 𝑥 ∙
𝐿 𝑥 = 2𝑥 + 5
𝐿′ 𝑥 = 2
𝐻′
𝑥 − 𝐻(𝑥) ∙ 𝐿′(𝑥)
=
𝐿(𝑥)2
1
2𝑥 + 5 ∙ − (ln 𝑥) ∙ 2
𝑥
𝑦′ =
2𝑥 + 5 2
𝑦′
ln 𝑥
2𝑥+5
2𝑥 + 5 − 2𝑥 ln 𝑥
𝑥
𝑦′ =
2𝑥 + 5 2
2𝑥 + 5 − 2𝑥ln 𝑥
𝑦 =
𝑥 2𝑥 + 5 2
′
2𝑥 + 5
− 2 ln 𝑥
𝑥
′
𝑦 =
2𝑥 + 5 2
Barnett/Ziegler/Byleen Business Calculus 12e
11
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
12
Objectives for Section 4.3
Day 2
The student will be able to :
■ Review derivatives of
products and quotients
𝐿𝐻′ − 𝐻𝐿′
𝑦′ =
𝐿2
■ Find the equations of tangent
lines
■ Solve applications
Barnett/Ziegler/Byleen Business Calculus 12e
13
Review
 What is the mnemonic for finding the derivative of a
product 𝑦 = 𝐿𝑅 ?
 𝑦 ′ = 𝐿𝑅′ + 𝑅𝐿′
 What is the mnemonic for finding the derivative of a
𝐻
quotient 𝑦 = ?
𝐿
′
 𝑦 =
𝐿𝐻 ′ −𝐻𝐿′
𝐿2
Barnett/Ziegler/Byleen Business Calculus 12e
14
Example 1
 Find ℎ′ 𝑥 , where f(x) is an unspecified differentiable
function.
ℎ 𝑥 = 2𝑥 4 𝑓(𝑥)
ℎ′ 𝑥 = 𝐿𝑅′ + 𝑅𝐿′
ℎ′ 𝑥 = 2𝑥 4 ∙ 𝑓 ′ 𝑥 + 𝑓(𝑥) ∙ 8𝑥 3
ℎ′ 𝑥 = 2𝑥 3 𝑥𝑓 ′ 𝑥 + 4𝑓(𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
Factor 2𝑥 3
from each
term.
15
Example 2
 Find ℎ′ 𝑥 , where f(x) is an unspecified differentiable
function.
′
𝐿𝐻
− 𝐻𝐿′
′
𝑓(𝑥)
ℎ 𝑥 =
2
ℎ 𝑥 = 3
𝐿
𝑥
3 𝑓′(𝑥) − 𝑓(𝑥)3𝑥 2
𝑥
ℎ′ 𝑥 =
𝑥3 2
Divide each
term by 𝑥 2 .
3 𝑓′(𝑥) − 𝑓(𝑥)3𝑥 2
𝑥
ℎ′ 𝑥 =
𝑥6
ℎ′
𝑥𝑓′(𝑥) − 3𝑓 𝑥
𝑥 =
𝑥4
Barnett/Ziegler/Byleen Business Calculus 12e
16
Example 3
Let f (x) = (x2 + 6)(ln x). Find the equation of the line tangent
to the graph of f (x) at x = 1.
1
𝑓 𝑥 = 𝑥 + 6 + ln 𝑥 2𝑥
𝑥
1
′
𝑓 1 = 1 + 6 + ln 1 2(1)
1
𝑓′ 1 = 7 + 0 ∙ 2 = 7
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑓 1 = (7)(ln 1) = 7 ∙ 0 = 0
𝑦 − 0 = 7(𝑥 − 1)
′
2
𝑚=7
(1,0)
𝑦 = 7𝑥 − 7
Barnett/Ziegler/Byleen Business Calculus 12e
17
Example 4
3𝑥
.
2𝑥−1
Let f (x) =
Find the equation of the line tangent to the
graph of f (x) at x = 2.
𝑓′
2𝑥 − 1 3 − 3𝑥(2)
𝑥 =
(2𝑥 − 1)2
𝑓′
−3
6𝑥 − 3 − 6𝑥
=
𝑥 =
(2𝑥 − 1)2
(2𝑥 − 1)2
−3 −1
𝑓 2 =
=
9
3
′
2, 𝑓 2
= (2,2)
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
−1
𝑦−2=
(𝑥 − 2)
3
−1
8
𝑦=
𝑥+
3
3
Barnett/Ziegler/Byleen Business Calculus 12e
18
Application 1
 The total sales (in thousands of games) of Call of Duty x
months after the game is introduced is given by:
150𝑥
𝑇 𝑥 =
𝑥+3
 A) Find 𝑇 ′ 𝑥
 B) Find 𝑇 12 and 𝑇 ′ 12 and interpret the results.
 C) Use the results from (B) to estimate the total sales after
13 months.
Barnett/Ziegler/Byleen Business Calculus 12e
19
150𝑥
𝑇 𝑥 =
𝑥+3
Application 1
(continued)
 A) Find 𝑇 ′ 𝑥
′ − 𝐻𝐿′
𝐿𝐻
𝑇′ 𝑥 =
𝐿2
𝑇′
𝑥 + 3 150 − 150𝑥(1)
𝑥 =
(𝑥 + 3)2
𝑇′
150𝑥 + 450 − 150𝑥
𝑥 =
(𝑥 + 3)2
𝑇′
450
𝑥 =
(𝑥 + 3)2
Barnett/Ziegler/Byleen Business Calculus 12e
20
Application 1
(continued)
 B) Find 𝑇 12 and 𝑇 ′ 12 and interpret the results
150𝑥
𝑇 𝑥 =
𝑥+3
𝑇 12 =
150(12)
12 + 3
𝑇′
450
𝑥 =
(𝑥 + 3)2
𝑇 ′ 12 =
450
(12 + 3)2
1800
𝑇 12 =
15
450
𝑇 12 =
225
𝑇 12 = 120
𝑇 ′ 12 = 2
′
After 12 months, the total sales are 120,000
games and sales are increasing at a rate of
2,000 games per month.
Barnett/Ziegler/Byleen Business Calculus 12e
21
Application 1
(continued)
 C) Use the results from (B) to estimate the total sales after
13 months.
𝑇 12 + 𝑇′(12)
120 + 2 = 122
After 13 months, the total sales will be 122,000
games.
Barnett/Ziegler/Byleen Business Calculus 12e
22
Homework
#4-3B: Pg 231
31, 35, 39, 43,
49, 51, 81, 83, 87
Barnett/Ziegler/Byleen Business Calculus 12e
23