Chapter 4
Additional
Derivative Topics
Section 4
The Chain Rule
Part I
Learning Objectives for Section
4.4
The Chain Rule
 The student will be able to identify the
composition of two functions.
 The student will be able to apply the
chain rule in order to evaluate more
complex derivatives.
Barnett/Ziegler/Byleen Business Calculus 12e
2
Composite Functions
Definition: A function m is a composite of functions f and g if
m(x) = f [g(x)]
The domain of m is the set of all numbers x such that x is in the
domain of g and g(x) is in the domain of f.
Barnett/Ziegler/Byleen Business Calculus 12e
3
Composite Functions
 Ex 1: 𝑚 𝑥 = 3𝑥 + 4 2
 We can express 𝑚 𝑥 as a composition of functions.
• 𝑔 𝑥 = 3𝑥 + 4
• 𝑓 𝑥 = 𝑥2
• 𝑚 𝑥 = 𝑓 𝑔(𝑥)
 Ex 2: 𝑚 𝑥 = 𝑒 4𝑥−1
 We can express 𝑚 𝑥 as a composition of functions.
• 𝑔 𝑥 = 4𝑥 − 1
• 𝑓 𝑥 = 𝑒𝑥
• 𝑚 𝑥 = 𝑓 𝑔(𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
4
Composite Functions
 Ex 3: Given 𝑚(𝑥), identify 𝑓(𝑥) and 𝑔(𝑥) such that
𝑚(𝑥) = 𝑓 𝑔(𝑥)
Note: There is more than one answer.
4
𝑚(𝑥) = ln 3
𝑥
4
𝑔 𝑥 = 3
𝑥
𝑚(𝑥) =
3
5 + 𝑥2
𝑔 𝑥 = 5 + 𝑥2
𝑓 𝑥 =
3
𝑥
𝑓 𝑥 = ln 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
5
Derivative Notation
 Recall the alternate notation for derivatives.
 y is a function of x:
′
• 𝑦=𝑓 𝑥 y =
𝑑𝑦
𝑑𝑥
 u is a function of x:
=
𝑑𝑢
𝑑𝑥
• 𝑦 = 𝑓(𝑢)  y =
𝑑𝑦
𝑑𝑢
• 𝑢 = 𝑓(𝑥) 
u′
 y is a function of u:
′
Barnett/Ziegler/Byleen Business Calculus 12e
6
Chain Rule
The chain rule enables us to compute the derivatives of
many composite functions of the form 𝑓 𝑔(𝑥) .
It is sometimes referred to as the “outside-inside” rule.
Chain Rule: If y = f (u) and u = g(x) define the
composite function y = f (u) = f [g(x)], then
dy dy du
dy
du


, provided
and
exist .
dx du dx
du
dx
Or, another way to look at it is:
𝑦 ′ = 𝑓′(𝑔 𝑥 ) ∙ 𝑔′(𝑥)
Barnett/Ziegler/Byleen Business Calculus 12e
7
Derivatives Using the Chain Rule
Ex. 4: 𝑦 = 2𝑥 3 + 1 5
𝑑𝑢
3
𝑢 = 2𝑥 + 1
= 6𝑥 2
𝑑𝑥
𝑑𝑦
= 5𝑢4
𝑦 = 𝑢5
𝑑𝑢
𝑑𝑦 𝑑𝑦 𝑑𝑢
𝑦 =
=
∙
𝑑𝑥 𝑑𝑢 𝑑𝑥
= 5𝑢4 6𝑥 2
′
=5
2𝑥 3
+1
Ex 5: 𝑦 = ln 𝑥 4
𝑑𝑢 1
𝑢 = ln 𝑥
=
𝑑𝑥 𝑥
𝑑𝑦
4
𝑦=𝑢
= 4𝑢3
𝑑𝑢
𝑦′
4 6𝑥 2
= 30𝑥 2 2𝑥 3 + 1
4
Barnett/Ziegler/Byleen Business Calculus 12e
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
=
∙
𝑑𝑥 𝑑𝑢 𝑑𝑥
= 4 ln 𝑥
4 ln 𝑥
=
𝑥
3
1
∙
𝑥
3
8
Derivatives Using the Chain Rule
Ex. 6: 𝑦 = 𝑒 𝑥 − 2𝑥 7
𝑑𝑢
𝑥
𝑢 = 𝑒 − 2𝑥
= 𝑒𝑥 − 2
𝑑𝑥
𝑑𝑦
7
𝑦=𝑢
= 7𝑢6
𝑑𝑢
𝑑𝑦 𝑑𝑢
𝑦′ =
∙
𝑑𝑢 𝑑𝑥
𝑑𝑦 𝑑𝑢
𝑦′ =
∙
𝑑𝑢 𝑑𝑥
= 7 𝑒 𝑥 − 2𝑥
Ex 7: 𝑦 = 5𝑒 −4𝑥
𝑑𝑢
𝑢 = −4𝑥
= −4
𝑑𝑥
𝑑𝑦
𝑢
= 5𝑒 𝑢
𝑦 = 5𝑒
𝑑𝑢
6
∙ 𝑒𝑥 − 2
Barnett/Ziegler/Byleen Business Calculus 12e
= 5𝑒 −4𝑥 (−4)
= −20𝑒 −4𝑥
9
Derivatives Using the Chain Rule
Ex. 8: 𝑦 =
𝑢=
2𝑥 3
𝑦=8
𝑦′
3 +4
2𝑥
8
+4
𝑑𝑦
𝑑𝑢
𝑢
𝑑𝑢
= 6𝑥 2
𝑑𝑥
Ex. 9: 𝑦 = log 3 5𝑥 2 + 1
= 8𝑢 ∙ ln 8
𝑑𝑦 𝑑𝑢
=
∙
𝑑𝑢 𝑑𝑥
=8
=
2𝑥 3 +4
6𝑥 2
∙
∙ ln 8 ∙ 6𝑥 2
3 +4
2𝑥
8
∙ ln 8
Barnett/Ziegler/Byleen Business Calculus 12e
𝑑𝑢
𝑢=
+1
= 10x
𝑑𝑥
𝑑𝑦
1
𝑦 = log 3 𝑢
=
𝑑𝑢 𝑢 ln 3
𝑑𝑦 𝑑𝑢
′
𝑦 =
∙
𝑑𝑢 𝑑𝑥
1
=
∙ 10𝑥
2
5𝑥 + 1 ln 3
5𝑥 2
10𝑥
=
(5𝑥 2 + 1)(ln 3)
10
Derivatives Using the Chain Rule
𝐸𝑥. 10
𝑦 = ln 3𝑥 2
1
′
𝑦 = 2 ∙ 6𝑥
3𝑥
6𝑥
′
𝑦 = 2
3𝑥
𝐸𝑥. 11
𝑦 = ln 𝑥 2 + 2𝑥 + 1
2
′
𝑦 =
𝑥
1
𝑦′ = 2
∙ 2𝑥 + 2
𝑥 + 2𝑥 + 1
2𝑥 + 2
= 2
𝑥 + 2𝑥 + 1
2(𝑥 + 1)
=
𝑥+1 𝑥+1
2
=
𝑥+1
Barnett/Ziegler/Byleen Business Calculus 12e
11
Derivatives Using the Chain Rule
𝐸𝑥. 12
1
𝑦= 2
𝑥 −7
𝑦 = 𝑥2 − 7
4
−4
𝑦′ = −4 𝑥 2 − 7
−5
= −8𝑥 𝑥 2 − 7
Barnett/Ziegler/Byleen Business Calculus 12e
2𝑥
−5
12
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
13
Chapter 4
Additional
Derivative Topics
Section 4
The Chain Rule
Massacre
(Part II)
Product Rule w/ Chain Rule
𝑓 𝑥 = 𝑥 2 3𝑥 − 4
5
𝑓 ′ 𝑥 = 𝐿𝑅′ + 𝑅𝐿′
L
R′
R
𝑓 ′ 𝑥 = 𝑥 2 ∙ 5 3𝑥 − 4 4 ∙ 3 + 3𝑥 − 4
𝑓 ′ 𝑥 = 15𝑥 2 3𝑥 − 4
4
+ 2𝑥 3𝑥 − 4
L′
5
∙ 2𝑥
5
(𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑜𝑢𝑡 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑡𝑒𝑟𝑚𝑠)
𝑓 ′ 𝑥 = 𝑥 3𝑥 − 4
4
15𝑥 + 2(3𝑥 − 4)
𝑓 ′ 𝑥 = 𝑥 3𝑥 − 4
4
15𝑥 + 6𝑥 − 8
𝑓 ′ 𝑥 = 𝑥 3𝑥 − 4
4
21𝑥 − 8
Barnett/Ziegler/Byleen Business Calculus 12e
15
Quotient Rule w/ Chain Rule
𝑥4
𝑓 𝑥 =
3𝑥 − 8
′
2
𝐻 = 𝑥4
𝐿 = 3𝑥 − 8
𝐿𝐻 − 𝐻𝐿′
𝑥 =
𝐿2
3𝑥 − 8 2 4𝑥 3 − 𝑥 4 6 3𝑥 − 8
𝑓′(𝑥) =
2
2
3𝑥 − 8
𝑓′
4𝑥 3 3𝑥 − 8 2 − 6𝑥 4 3𝑥 − 8
=
3𝑥 − 8 4
2𝑥 3 (3𝑥 − 8) 2 3𝑥 − 8 − 3𝑥
=
3𝑥 − 8 4
Barnett/Ziegler/Byleen Business Calculus 12e
𝐻′ = 4𝑥 3
2
𝐿′ = 2 3𝑥 − 8 1 3
𝐿′ = 6 3𝑥 − 8
2𝑥 3 3𝑥 − 16
=
3𝑥 − 8 3
16
Equation of Tangent Line
𝑓 𝑥 = 2𝑥 + 8
1 2
Find the equation of the line tangent
to the graph of f when x=4.
1
𝑓′ 𝑥 = 2𝑥 + 8 −1 2 2
2
= 2𝑥 + 8 −1 2
𝑓 4 =4
𝑓′
1
4 =
4
Barnett/Ziegler/Byleen Business Calculus 12e
4,4 𝑚 = 14
𝑦 − 4 = 14
𝑥−4
𝑦 − 4 = 14𝑥−1
𝑦 = 14𝑥+3
17
Horizontal Tangent Line
𝑥
𝑓 𝑥 =
2𝑥 + 5
2
Find the value(s) of x where the
tangent line is horizontal.
′
𝐻=𝑥
𝐻′ = 1
𝐿𝐻
−
𝐻𝐿′
𝑓′ 𝑥 =
𝐿 = 2𝑥 + 5 2 𝐿′ = 2 2𝑥 + 5 1 2
𝐿2
2𝑥 + 5 2 (1) − (𝑥)4 2𝑥 + 5
𝐿′ = 4 2𝑥 + 5
𝑓′(𝑥) =
2
2
2𝑥 + 5
2𝑥 + 5
2
− 4𝑥 2𝑥 + 5
=
2𝑥 + 5 4
(2𝑥 + 5) 2𝑥 + 5 − 4𝑥
=
2𝑥 + 5 4
Barnett/Ziegler/Byleen Business Calculus 12e
5 − 2𝑥
=
2𝑥 + 5 3
18
Continued
5 − 2𝑥
𝑓′(𝑥) =
2𝑥 + 5 3
The tangent line is horizontal when
its slope is zero.
5 − 2𝑥
0=
2𝑥 + 5 3
0 = 5 − 2𝑥
2𝑥 = 5
𝑥 = 2.5
The tangent line is horizontal when
x=2.5
Barnett/Ziegler/Byleen Business Calculus 12e
19
Application
 The total cost (in hundreds of dollars) of producing x
cameras per week is: 𝐶 𝑥 = 6 + 4𝑥 + 4 , 0 ≤ 𝑥 ≤ 30
 A) Find 𝐶 ′ 𝑥
 B) Find 𝐶 15 𝑎𝑛𝑑 𝐶′(15) and interpret the results.
𝐶 ′ 𝑥 = 12
𝐶′
𝑥 =
4𝑥+4 −1 2 4
2 4𝑥+4 −1 2
𝐶 15 = 6 + 4 ∙ 15 + 4 = 14
𝐶 ′ 15 = 2 4 ∙ 15 + 4 −1
2
1
′
𝐶 15 =
= = .25
64 4
2
At a production level of 15 cameras per week, the total cost is
$1400 and it is increasing at a rate of $25 per camera.
Barnett/Ziegler/Byleen Business Calculus 12e
20
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
21