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Chapter 9
Trigonometric
Functions
Section 2
Derivatives of
Trigonometric
Functions
Part 1
Learning Objectives for Section 9.2
Derivatives of Trig Functions
■ The student will be able to
use and apply the formulas
for derivatives of the
trigonometric functions.
Barnett/Ziegler/Byleen Business Calculus 12e
2
Review of Derivatives
 Power Rule w/ Chain Rule:
• 𝑦 = 𝑓(𝑥) 𝑛
𝑦 ′ = 𝑛 𝑓(𝑥)
 Natural Log w/ Chain Rule:
• 𝑦 = ln 𝑓(𝑥)
𝑦′
=
𝑛−1 𝑓′(𝑥)
1
𝑓′(𝑥)
𝑓(𝑥)
 Exponential w/ Chain Rule:
• 𝑦 = 𝑒 𝑓(𝑥)
𝑦 ′ = 𝑒 𝑓(𝑥) 𝑓′(𝑥)
 Product Rule:
• 𝑦 = 𝐿 𝑥 𝑅(𝑥) 𝑦 ′ = 𝐿 𝑥 𝑅′ 𝑥 + 𝑅 𝑥 𝐿′(𝑥)
 Quotient Rule:
• 𝑦=
𝐻(𝑥)
𝐿(𝑥)
𝑦′
=
𝐿 𝑥 𝐻 ′ 𝑥 −𝐻 𝑥 𝐿′ 𝑥
𝐿(𝑥) 2
Barnett/Ziegler/Byleen Business Calculus 12e
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Derivative Formulas
For Sine and Cosine
Basic Form:
𝑦 = sin 𝑥
𝑦 ′ = cos 𝑥
𝑦 = cos 𝑥
𝑦 ′ = − sin 𝑥
Chain Rule:
𝑦 = sin 𝑓(𝑥)
𝑦 ′ = cos 𝑓(𝑥) ∙ 𝑓′(𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
𝑦 = cos 𝑓(𝑥)
𝑦 ′ = − sin 𝑓(𝑥) ∙ 𝑓′(𝑥)
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Example 1
Find the derivative of each function given.
𝑦 = 2 sin 𝑥
𝑦 = sin(3𝑥)
𝑦 = sin(𝑥 2 )
𝑦′ = 2 cos 𝑥
𝑦′ = cos(3𝑥) ∙ 3
𝑦′ = 𝑐𝑜𝑠(𝑥 2 ) ∙ 2𝑥
𝑦′ = 3 cos(3𝑥)
𝑦′ = 2x 𝑐𝑜𝑠(𝑥 2 )
𝑦 = sin 𝑥
3
𝑦′ = 3 sin 𝑥
𝑦 = sin 4𝑥
2
∙ cos 𝑥
𝑦′ = 3 cos 𝑥 𝑠𝑖𝑛2 𝑥
3
𝑦′ = 3 sin 4𝑥 2 ∙ (cos 4𝑥) ∙ 4
𝑦′ = 12 sin 4𝑥 2 ∙ (cos 4𝑥)
𝑦′ = 12 𝑠𝑖𝑛2 4𝑥 ∙ (cos 4𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 2
Find the derivative of each function given.
𝑦 = 5 𝑐𝑜𝑠 𝑥
𝑦 ′ = −5 𝑠𝑖𝑛 𝑥
𝑦 = 𝑐𝑜𝑠 𝑥
𝑦 = cos(8𝑥)
𝑦 = 𝑐𝑜𝑠(𝑥 3 )
𝑦′ = −𝑠𝑖𝑛(8𝑥) ∙ 8
𝑦′ = −𝑠𝑖𝑛(𝑥 3 ) ∙ 3𝑥 2
𝑦 ′ = −8 𝑠𝑖𝑛(8𝑥)
𝑦′ = −3𝑥 2 𝑠𝑖𝑛 (𝑥 3 )
4
𝑦′ = 4 𝑐𝑜𝑠 𝑥
3 (−𝑠𝑖𝑛 𝑥)
𝑦 ′ = −4 𝑐𝑜𝑠 𝑥
3 (𝑠𝑖𝑛 𝑥)
𝑦 ′ = −4(𝑐𝑜𝑠 3 𝑥) (𝑠𝑖𝑛 𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3
Find the derivative of tan x:
sin 𝑥
𝑦 = tan 𝑥 =
cos 𝑥
′
𝐿𝐻
− 𝐻𝐿′
′
𝑦 =
𝐿2
cos 𝑥 cos 𝑥 − sin 𝑥 (− sin 𝑥)
′
𝑦 =
cos 𝑥 2
2 𝑥 + 𝑠𝑖𝑛2 𝑥
𝑐𝑜𝑠
𝑦′ =
𝑐𝑜𝑠 2 𝑥
1
=
𝑐𝑜𝑠 2 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦:
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 = 1
= 𝑠𝑒𝑐 2 𝑥
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Example 4
Find the derivative of the function given.
1
𝑦 = 𝑠𝑒𝑐 𝑥 =
cos 𝑥
𝑦 ′ = cos 𝑥
−1
𝑦 ′ = −1 cos 𝑥 −2 ∙ (− sin 𝑥)
sin 𝑥
1
sin 𝑥
sin 𝑥
′
= tan 𝑥 sec 𝑥
=
∙
𝑦 =
=
cos 𝑥 cos 𝑥
cos 𝑥 2
cos 𝑥 cos 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 5
Find the derivative of the function given.
𝑦 = sin(𝑥𝑒 𝑥 )
𝑦 ′ = cos(𝑥𝑒 𝑥 ) ∙ (𝐿𝑅′ + 𝑅𝐿′ )
𝑦 ′ = cos(𝑥𝑒 𝑥 ) ∙ (𝑥𝑒 𝑥 + 𝑒 𝑥 ∙ 1)
𝑦 ′ = cos(𝑥𝑒 𝑥 )𝑒 𝑥 𝑥 + 1
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 6
Find the derivative of each function given.
𝑦 = ln(cos 𝑥)
1
′
𝑦 =
∙ (− sin 𝑥)
cos 𝑥
𝑦 = 𝑒 sin 𝑥
𝑦 ′ = 𝑒 sin 𝑥 ∙ cos 𝑥
𝑦 ′ = − tan 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
10
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
11
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
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Chapter 9
Trigonometric
Functions
Section 2
Derivatives of
Trigonometric
Functions
Part 2
Learning Objectives for Section 9.2
Derivatives of Trig Functions
■ The student will be able to
solve applications of
derivatives of trigonometric
functions.
Barnett/Ziegler/Byleen Business Calculus 12e
14
The Meaning of Derivatives
 Recall the geometric interpretation of a derivative:
• The derivative of a function is a “slope machine”.
• When you “plug” an x value into 𝑓 ′ 𝑥 , it gives you a
slope.
• It is the slope of the line tangent to the graph of f(x) at
the point (x, f(x)).
Barnett/Ziegler/Byleen Business Calculus 12e
15
Example 1: Slope
A) Find the slope of the graph of f (x) = sin x at 𝑥 =
𝜋
.
2
𝜋
2
B) Write the equation of the line tangent to f(x) at 𝑥 = .
𝑓 ′ 𝑥 = cos 𝑥
𝜋
′
𝑓
= cos 𝜋2
2
=0
𝑆𝑙𝑜𝑝𝑒 = 0
𝑓 𝑥 = sin 𝑥
𝜋
𝑓
=1
2
𝜋
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑝𝑎𝑠𝑠𝑒𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ
,1
2
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒: 𝑦 − 1 = 0 𝑥 − 𝜋2
𝑦=1
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 2: Slope
A) Find the slope of the graph of 𝑓 𝑥 = cos 2𝑥 𝑎𝑡 𝑥 =
B) Write the equation of the tangent line in point-slope
form.
𝑓 𝑥 = cos(2𝑥)
𝑓 ′ 𝑥 = −2sin(2𝑥)
𝑓 𝜋6 = 𝑐𝑜𝑠 𝜋3
𝑓 ′ 𝜋6 = −2sin 𝜋3
= 12
= −2
3
2
𝑆𝑙𝑜𝑝𝑒 = − 3
𝜋
6
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑝𝑎𝑠𝑠𝑒𝑠
𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝜋6, 12
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒: 𝑦 − 12 = − 3 𝑥 − 𝜋6
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3: Slope
A) Find the slope of the graph of 𝑓 𝑥 = sin 𝑥 2 𝑎𝑡 𝑥 =
B) Write the equation of the tangent line in point-slope
form.
𝑓 𝑥 = sin 𝑥 2
𝑓 ′ 𝑥 = 2(sin 𝑥)(cos 𝑥)
2
𝜋
2
𝑓 4 = 2
𝜋
2
2
′
𝑓 4 =2 2
2
= 12
=1
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑝𝑎𝑠𝑠𝑒𝑠
𝜋 1
𝑡ℎ𝑟𝑜𝑢𝑔ℎ
𝑆𝑙𝑜𝑝𝑒 = 1
4, 2
𝜋
4
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒: 𝑦 − 12 = 𝑥 − 𝜋4
Barnett/Ziegler/Byleen Business Calculus 12e
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More Derivative Applications
 The derivative also represents the rate of change of f(x)
with respect to x.
 For example:
• 𝑅(𝑥) = the total revenue generated at a production level
of x products
• 𝑅′(𝑥) = the rate at which the revenue is
increasing/decreasing at a production level of x
products. (also known as marginal revenue)
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 4: Application
A soft-drink company has revenues from sales over a 2-year period
as given approximately by
R(t) = 4 − 3𝑐𝑜𝑠 𝜋𝑡
0  t  24
6
where R(t) is revenue (in millions of dollars) for a month of sales t
months after February 1.
a. What is the total revenue 6 months after February 1?
b. What is the rate of change of revenue t months after February 1?
c. What is the rate of change of revenue 6 months after February 1?
d. What is the rate of change of revenue 8 months after February 1?
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 4: Application
A soft-drink company has revenues from sales over a 2-year period
as given approximately by
R(t) = 4 − 3𝑐𝑜𝑠 𝜋𝑡
0  t  24
6
where R(t) is revenue (in millions of dollars) for a month of sales t
months after February 1.
a. What is the total revenue 6 months after February 1?
𝑅 6 = 4 − 3𝑐𝑜𝑠 𝜋
= 4 − 3(−1)
=7
The total revenue is $7 million for a month of sales
6 months after Feb. 1.
Barnett/Ziegler/Byleen Business Calculus 12e
21
Example 4: Application
A soft-drink company has revenues from sales over a 2-year period
as given approximately by
R(t) = 4 − 3𝑐𝑜𝑠 𝜋𝑡
0  t  24
6
where R(t) is revenue (in millions of dollars) for a month of sales t
months after February 1.
b. What is the rate of change of revenue t months after February 1?
𝑅′ 𝑡 = −3 − sin 𝜋𝑡
6 ∙
𝑅′ 𝑡 = 𝜋2 sin
𝜋
6
𝜋𝑡
6
Barnett/Ziegler/Byleen Business Calculus 12e
22
Example 4: Application
A soft-drink company has revenues from sales over a 2-year period
as given approximately by
R(t) = 4 − 3𝑐𝑜𝑠 𝜋𝑡
0  t  24
6
where R(t) is revenue (in millions of dollars) for a month of sales t
months after February 1.
c. What is the rate of change of revenue 6 months after February 1?
𝑅′ 𝑡 = 𝜋2 sin
𝜋𝑡
6
𝑅′ 6 = 𝜋2 sin
𝜋
=0
The revenue isn’t increasing or decreasing 6
months after Feb. 1.
Barnett/Ziegler/Byleen Business Calculus 12e
23
Example 4: Application
A soft-drink company has revenues from sales over a 2-year period
as given approximately by
R(t) = 4 − 3𝑐𝑜𝑠 𝜋𝑡
0  t  24
6
where R(t) is revenue (in millions of dollars) for a month of sales t
months after February 1.
d. What is the rate of change of revenue 8 months after February 1?
𝑅′ 𝑡 = 𝜋2 sin
𝑅′ 8 = 𝜋2 sin
𝜋𝑡
6
4𝜋
3
= 𝜋2
− 3
2
≈ −1.4
The revenue is decreasing at a rate of $1.4 million
per month 8 months after Feb. 1.
Barnett/Ziegler/Byleen Business Calculus 12e
24
Example 4: Application
To verify the correctness of our results, let’s look at the
graph of R(x).
When t=6, the slope is 0.
Revenue is not increasing
or decreasing.
Barnett/Ziegler/Byleen Business Calculus 12e
When t=8, the slope is -1.4.
Revenue is decreasing.
25
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
26