Calculus 4.4 power point lesson

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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: π‘š π‘₯ = 3π‘₯ + 4 2
 We can express π‘š π‘₯ as a composition of functions.
β€’ 𝑔 π‘₯ = 3π‘₯ + 4
β€’ 𝑓 π‘₯ = π‘₯2
β€’ π‘š π‘₯ = 𝑓 𝑔(π‘₯)
 Ex 2: π‘š π‘₯ = 𝑒 4π‘₯βˆ’1
 We can express π‘š π‘₯ as a composition of functions.
β€’ 𝑔 π‘₯ = 4π‘₯ βˆ’ 1
β€’ 𝑓 π‘₯ = 𝑒π‘₯
β€’ π‘š π‘₯ = 𝑓 𝑔(π‘₯)
Barnett/Ziegler/Byleen Business Calculus 12e
4
Composite Functions
 Ex 3: Given π‘š(π‘₯), identify 𝑓(π‘₯) and 𝑔(π‘₯) such that
π‘š(π‘₯) = 𝑓 𝑔(π‘₯)
Note: There is more than one answer.
4
π‘š(π‘₯) = ln 3
π‘₯
4
𝑔 π‘₯ = 3
π‘₯
π‘š(π‘₯) =
3
5 + π‘₯2
𝑔 π‘₯ = 5 + π‘₯2
𝑓 π‘₯ =
3
π‘₯
𝑓 π‘₯ = ln π‘₯
Barnett/Ziegler/Byleen Business Calculus 12e
5
Derivative Notation
 Recall the alternate notation for derivatives.
 y is a function of x:
β€²
β€’ 𝑦=𝑓 π‘₯ οƒ y =
𝑑𝑦
𝑑π‘₯
 u is a function of x:
=
𝑑𝑒
𝑑π‘₯
β€’ 𝑦 = 𝑓(𝑒) οƒ  y =
𝑑𝑦
𝑑𝑒
β€’ 𝑒 = 𝑓(π‘₯) οƒ 
uβ€²
 y is a function of u:
β€²
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(π‘₯) .
It is sometimes referred to as the β€œoutside-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
ο€½
οƒ—
, provided
and
exist .
dx du dx
du
dx
Or, another way to look at it is:
𝑦 β€² = 𝑓′(𝑔 π‘₯ ) βˆ™ 𝑔′(π‘₯)
Barnett/Ziegler/Byleen Business Calculus 12e
7
Derivatives Using the Chain Rule
Ex. 4: 𝑦 = 2π‘₯ 3 + 1 5
𝑑𝑒
3
𝑒 = 2π‘₯ + 1
= 6π‘₯ 2
𝑑π‘₯
𝑑𝑦
= 5𝑒4
𝑦 = 𝑒5
𝑑𝑒
𝑑𝑦 𝑑𝑦 𝑑𝑒
𝑦 =
=
βˆ™
𝑑π‘₯ 𝑑𝑒 𝑑π‘₯
= 5𝑒4 6π‘₯ 2
β€²
=5
2π‘₯ 3
+1
Ex 5: 𝑦 = ln π‘₯ 4
𝑑𝑒 1
𝑒 = ln π‘₯
=
𝑑π‘₯ π‘₯
𝑑𝑦
4
𝑦=𝑒
= 4𝑒3
𝑑𝑒
𝑦′
4 6π‘₯ 2
= 30π‘₯ 2 2π‘₯ 3 + 1
4
Barnett/Ziegler/Byleen Business Calculus 12e
𝑑𝑦 𝑑𝑦 𝑑𝑒
=
=
βˆ™
𝑑π‘₯ 𝑑𝑒 𝑑π‘₯
= 4 ln π‘₯
4 ln π‘₯
=
π‘₯
3
1
βˆ™
π‘₯
3
8
Derivatives Using the Chain Rule
Ex. 6: 𝑦 = 𝑒 π‘₯ βˆ’ 2π‘₯ 7
𝑑𝑒
π‘₯
𝑒 = 𝑒 βˆ’ 2π‘₯
= 𝑒π‘₯ βˆ’ 2
𝑑π‘₯
𝑑𝑦
7
𝑦=𝑒
= 7𝑒6
𝑑𝑒
𝑑𝑦 𝑑𝑒
𝑦′ =
βˆ™
𝑑𝑒 𝑑π‘₯
𝑑𝑦 𝑑𝑒
𝑦′ =
βˆ™
𝑑𝑒 𝑑π‘₯
= 7 𝑒 π‘₯ βˆ’ 2π‘₯
Ex 7: 𝑦 = 5𝑒 βˆ’4π‘₯
𝑑𝑒
𝑒 = βˆ’4π‘₯
= βˆ’4
𝑑π‘₯
𝑑𝑦
𝑒
= 5𝑒 𝑒
𝑦 = 5𝑒
𝑑𝑒
6
βˆ™ 𝑒π‘₯ βˆ’ 2
Barnett/Ziegler/Byleen Business Calculus 12e
= 5𝑒 βˆ’4π‘₯ (βˆ’4)
= βˆ’20𝑒 βˆ’4π‘₯
9
Derivatives Using the Chain Rule
Ex. 8: 𝑦 =
𝑒=
2π‘₯ 3
𝑦=8
𝑦′
3 +4
2π‘₯
8
+4
𝑑𝑦
𝑑𝑒
𝑒
𝑑𝑒
= 6π‘₯ 2
𝑑π‘₯
Ex. 9: 𝑦 = log 3 5π‘₯ 2 + 1
= 8𝑒 βˆ™ ln 8
𝑑𝑦 𝑑𝑒
=
βˆ™
𝑑𝑒 𝑑π‘₯
=8
=
2π‘₯ 3 +4
6π‘₯ 2
βˆ™
βˆ™ ln 8 βˆ™ 6π‘₯ 2
3 +4
2π‘₯
8
βˆ™ ln 8
Barnett/Ziegler/Byleen Business Calculus 12e
𝑑𝑒
𝑒=
+1
= 10x
𝑑π‘₯
𝑑𝑦
1
𝑦 = log 3 𝑒
=
𝑑𝑒 𝑒 ln 3
𝑑𝑦 𝑑𝑒
β€²
𝑦 =
βˆ™
𝑑𝑒 𝑑π‘₯
1
=
βˆ™ 10π‘₯
2
5π‘₯ + 1 ln 3
5π‘₯ 2
10π‘₯
=
(5π‘₯ 2 + 1)(ln 3)
10
Derivatives Using the Chain Rule
𝐸π‘₯. 10
𝑦 = ln 3π‘₯ 2
1
β€²
𝑦 = 2 βˆ™ 6π‘₯
3π‘₯
6π‘₯
β€²
𝑦 = 2
3π‘₯
𝐸π‘₯. 11
𝑦 = ln π‘₯ 2 + 2π‘₯ + 1
2
β€²
𝑦 =
π‘₯
1
𝑦′ = 2
βˆ™ 2π‘₯ + 2
π‘₯ + 2π‘₯ + 1
2π‘₯ + 2
= 2
π‘₯ + 2π‘₯ + 1
2(π‘₯ + 1)
=
π‘₯+1 π‘₯+1
2
=
π‘₯+1
Barnett/Ziegler/Byleen Business Calculus 12e
11
Derivatives Using the Chain Rule
𝐸π‘₯. 12
1
𝑦= 2
π‘₯ βˆ’7
𝑦 = π‘₯2 βˆ’ 7
4
βˆ’4
𝑦′ = βˆ’4 π‘₯ 2 βˆ’ 7
βˆ’5
= βˆ’8π‘₯ π‘₯ 2 βˆ’ 7
Barnett/Ziegler/Byleen Business Calculus 12e
2π‘₯
βˆ’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 π‘₯ = π‘₯ 2 3π‘₯ βˆ’ 4
5
𝑓 β€² π‘₯ = 𝐿𝑅′ + 𝑅𝐿′
L
Rβ€²
R
𝑓 β€² π‘₯ = π‘₯ 2 βˆ™ 5 3π‘₯ βˆ’ 4 4 βˆ™ 3 + 3π‘₯ βˆ’ 4
𝑓 β€² π‘₯ = 15π‘₯ 2 3π‘₯ βˆ’ 4
4
+ 2π‘₯ 3π‘₯ βˆ’ 4
Lβ€²
5
βˆ™ 2π‘₯
5
(π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘’π‘‘ π‘œπ‘’π‘‘ π‘œπ‘“ π‘π‘œπ‘‘β„Ž π‘‘π‘’π‘Ÿπ‘šπ‘ )
𝑓 β€² π‘₯ = π‘₯ 3π‘₯ βˆ’ 4
4
15π‘₯ + 2(3π‘₯ βˆ’ 4)
𝑓 β€² π‘₯ = π‘₯ 3π‘₯ βˆ’ 4
4
15π‘₯ + 6π‘₯ βˆ’ 8
𝑓 β€² π‘₯ = π‘₯ 3π‘₯ βˆ’ 4
4
21π‘₯ βˆ’ 8
Barnett/Ziegler/Byleen Business Calculus 12e
15
Quotient Rule w/ Chain Rule
π‘₯4
𝑓 π‘₯ =
3π‘₯ βˆ’ 8
β€²
2
𝐻 = π‘₯4
𝐿 = 3π‘₯ βˆ’ 8
𝐿𝐻 βˆ’ 𝐻𝐿′
π‘₯ =
𝐿2
3π‘₯ βˆ’ 8 2 4π‘₯ 3 βˆ’ π‘₯ 4 6 3π‘₯ βˆ’ 8
𝑓′(π‘₯) =
2
2
3π‘₯ βˆ’ 8
𝑓′
4π‘₯ 3 3π‘₯ βˆ’ 8 2 βˆ’ 6π‘₯ 4 3π‘₯ βˆ’ 8
=
3π‘₯ βˆ’ 8 4
2π‘₯ 3 (3π‘₯ βˆ’ 8) 2 3π‘₯ βˆ’ 8 βˆ’ 3π‘₯
=
3π‘₯ βˆ’ 8 4
Barnett/Ziegler/Byleen Business Calculus 12e
𝐻′ = 4π‘₯ 3
2
𝐿′ = 2 3π‘₯ βˆ’ 8 1 3
𝐿′ = 6 3π‘₯ βˆ’ 8
2π‘₯ 3 3π‘₯ βˆ’ 16
=
3π‘₯ βˆ’ 8 3
16
Equation of Tangent Line
𝑓 π‘₯ = 2π‘₯ + 8
1 2
Find the equation of the line tangent
to the graph of f when x=4.
1
𝑓′ π‘₯ = 2π‘₯ + 8 βˆ’1 2 2
2
= 2π‘₯ + 8 βˆ’1 2
𝑓 4 =4
𝑓′
1
4 =
4
Barnett/Ziegler/Byleen Business Calculus 12e
4,4 π‘š = 14
𝑦 βˆ’ 4 = 14
π‘₯βˆ’4
𝑦 βˆ’ 4 = 14π‘₯βˆ’1
𝑦 = 14π‘₯+3
17
Horizontal Tangent Line
π‘₯
𝑓 π‘₯ =
2π‘₯ + 5
2
Find the value(s) of x where the
tangent line is horizontal.
β€²
𝐻=π‘₯
𝐻′ = 1
𝐿𝐻
βˆ’
𝐻𝐿′
𝑓′ π‘₯ =
𝐿 = 2π‘₯ + 5 2 𝐿′ = 2 2π‘₯ + 5 1 2
𝐿2
2π‘₯ + 5 2 (1) βˆ’ (π‘₯)4 2π‘₯ + 5
𝐿′ = 4 2π‘₯ + 5
𝑓′(π‘₯) =
2
2
2π‘₯ + 5
2π‘₯ + 5
2
βˆ’ 4π‘₯ 2π‘₯ + 5
=
2π‘₯ + 5 4
(2π‘₯ + 5) 2π‘₯ + 5 βˆ’ 4π‘₯
=
2π‘₯ + 5 4
Barnett/Ziegler/Byleen Business Calculus 12e
5 βˆ’ 2π‘₯
=
2π‘₯ + 5 3
18
Continued
5 βˆ’ 2π‘₯
𝑓′(π‘₯) =
2π‘₯ + 5 3
The tangent line is horizontal when
its slope is zero.
5 βˆ’ 2π‘₯
0=
2π‘₯ + 5 3
0 = 5 βˆ’ 2π‘₯
2π‘₯ = 5
π‘₯ = 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: 𝐢 π‘₯ = 6 + 4π‘₯ + 4 , 0 ≀ π‘₯ ≀ 30
 A) Find 𝐢 β€² π‘₯
 B) Find 𝐢 15 π‘Žπ‘›π‘‘ 𝐢′(15) and interpret the results.
𝐢 β€² π‘₯ = 12
𝐢′
π‘₯ =
4π‘₯+4 βˆ’1 2 4
2 4π‘₯+4 βˆ’1 2
𝐢 15 = 6 + 4 βˆ™ 15 + 4 = 14
𝐢 β€² 15 = 2 4 βˆ™ 15 + 4 βˆ’1
2
1
β€²
𝐢 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
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