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 ο§ The student will be able to identify the composition of two functions. ο§ 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 ο§ Ex 1: π π₯ = 3π₯ + 4 2 ο§ We can express π π₯ as a composition of functions. • π π₯ = 3π₯ + 4 • π π₯ = π₯2 • π π₯ = π π(π₯) ο§ Ex 2: π π₯ = π 4π₯−1 ο§ We can express π π₯ as a composition of functions. • π π₯ = 4π₯ − 1 • π π₯ = ππ₯ • π π₯ = π π(π₯) Barnett/Ziegler/Byleen Business Calculus 12e 4 Composite Functions ο§ Ex 3: Given π(π₯), identify π(π₯) and π(π₯) such that π(π₯) = π π(π₯) Note: There is more than one answer. 4 π(π₯) = ln 3 π₯ 4 π π₯ = 3 π₯ π(π₯) = 3 5 + π₯2 π π₯ = 5 + π₯2 π π₯ = 3 π₯ π π₯ = ln π₯ Barnett/Ziegler/Byleen Business Calculus 12e 5 Derivative Notation ο§ Recall the alternate notation for derivatives. ο§ y is a function of x: ′ • π¦=π π₯ ο y = ππ¦ ππ₯ ο§ u is a function of x: = ππ’ ππ₯ • π¦ = π(π’) ο y = ππ¦ ππ’ • π’ = π(π₯) ο u′ ο§ 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 π π(π₯) . 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 ο½ ο , provided and exist . dx du dx du dx Or, another way to look at it is: π¦ ′ = π′(π π₯ ) β π′(π₯) Barnett/Ziegler/Byleen Business Calculus 12e 7 Derivatives Using the Chain Rule Ex. 4: π¦ = 2π₯ 3 + 1 5 ππ’ 3 π’ = 2π₯ + 1 = 6π₯ 2 ππ₯ ππ¦ = 5π’4 π¦ = π’5 ππ’ ππ¦ ππ¦ ππ’ π¦ = = β ππ₯ ππ’ ππ₯ = 5π’4 6π₯ 2 ′ =5 2π₯ 3 +1 Ex 5: π¦ = ln π₯ 4 ππ’ 1 π’ = ln π₯ = ππ₯ π₯ ππ¦ 4 π¦=π’ = 4π’3 ππ’ π¦′ 4 6π₯ 2 = 30π₯ 2 2π₯ 3 + 1 4 Barnett/Ziegler/Byleen Business Calculus 12e ππ¦ ππ¦ ππ’ = = β ππ₯ ππ’ ππ₯ = 4 ln π₯ 4 ln π₯ = π₯ 3 1 β π₯ 3 8 Derivatives Using the Chain Rule Ex. 6: π¦ = π π₯ − 2π₯ 7 ππ’ π₯ π’ = π − 2π₯ = ππ₯ − 2 ππ₯ ππ¦ 7 π¦=π’ = 7π’6 ππ’ ππ¦ ππ’ π¦′ = β ππ’ ππ₯ ππ¦ ππ’ π¦′ = β ππ’ ππ₯ = 7 π π₯ − 2π₯ Ex 7: π¦ = 5π −4π₯ ππ’ π’ = −4π₯ = −4 ππ₯ ππ¦ π’ = 5π π’ π¦ = 5π ππ’ 6 β ππ₯ − 2 Barnett/Ziegler/Byleen Business Calculus 12e = 5π −4π₯ (−4) = −20π −4π₯ 9 Derivatives Using the Chain Rule Ex. 8: π¦ = π’= 2π₯ 3 π¦=8 π¦′ 3 +4 2π₯ 8 +4 ππ¦ ππ’ π’ ππ’ = 6π₯ 2 ππ₯ Ex. 9: π¦ = log 3 5π₯ 2 + 1 = 8π’ β ln 8 ππ¦ ππ’ = β ππ’ ππ₯ =8 = 2π₯ 3 +4 6π₯ 2 β β ln 8 β 6π₯ 2 3 +4 2π₯ 8 β ln 8 Barnett/Ziegler/Byleen Business Calculus 12e ππ’ π’= +1 = 10x ππ₯ ππ¦ 1 π¦ = log 3 π’ = ππ’ π’ ln 3 ππ¦ ππ’ ′ π¦ = β ππ’ ππ₯ 1 = β 10π₯ 2 5π₯ + 1 ln 3 5π₯ 2 10π₯ = (5π₯ 2 + 1)(ln 3) 10 Derivatives Using the Chain Rule πΈπ₯. 10 π¦ = ln 3π₯ 2 1 ′ π¦ = 2 β 6π₯ 3π₯ 6π₯ ′ π¦ = 2 3π₯ πΈπ₯. 11 π¦ = ln π₯ 2 + 2π₯ + 1 2 ′ π¦ = π₯ 1 π¦′ = 2 β 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 πΈπ₯. 12 1 π¦= 2 π₯ −7 π¦ = π₯2 − 7 4 −4 π¦′ = −4 π₯ 2 − 7 −5 = −8π₯ π₯ 2 − 7 Barnett/Ziegler/Byleen Business Calculus 12e 2π₯ −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 π π₯ = π₯ 2 3π₯ − 4 5 π ′ π₯ = πΏπ ′ + π πΏ′ L R′ R π ′ π₯ = π₯ 2 β 5 3π₯ − 4 4 β 3 + 3π₯ − 4 π ′ π₯ = 15π₯ 2 3π₯ − 4 4 + 2π₯ 3π₯ − 4 L′ 5 β 2π₯ 5 (ππππ‘ππππ ππ’π‘ ππ πππ‘β π‘ππππ ) π ′ π₯ = π₯ 3π₯ − 4 4 15π₯ + 2(3π₯ − 4) π ′ π₯ = π₯ 3π₯ − 4 4 15π₯ + 6π₯ − 8 π ′ π₯ = π₯ 3π₯ − 4 4 21π₯ − 8 Barnett/Ziegler/Byleen Business Calculus 12e 15 Quotient Rule w/ Chain Rule π₯4 π π₯ = 3π₯ − 8 ′ 2 π» = π₯4 πΏ = 3π₯ − 8 πΏπ» − π»πΏ′ π₯ = πΏ2 3π₯ − 8 2 4π₯ 3 − π₯ 4 6 3π₯ − 8 π′(π₯) = 2 2 3π₯ − 8 π′ 4π₯ 3 3π₯ − 8 2 − 6π₯ 4 3π₯ − 8 = 3π₯ − 8 4 2π₯ 3 (3π₯ − 8) 2 3π₯ − 8 − 3π₯ = 3π₯ − 8 4 Barnett/Ziegler/Byleen Business Calculus 12e π»′ = 4π₯ 3 2 πΏ′ = 2 3π₯ − 8 1 3 πΏ′ = 6 3π₯ − 8 2π₯ 3 3π₯ − 16 = 3π₯ − 8 3 16 Equation of Tangent Line π π₯ = 2π₯ + 8 1 2 Find the equation of the line tangent to the graph of f when x=4. 1 π′ π₯ = 2π₯ + 8 −1 2 2 2 = 2π₯ + 8 −1 2 π 4 =4 π′ 1 4 = 4 Barnett/Ziegler/Byleen Business Calculus 12e 4,4 π = 14 π¦ − 4 = 14 π₯−4 π¦ − 4 = 14π₯−1 π¦ = 14π₯+3 17 Horizontal Tangent Line π₯ π π₯ = 2π₯ + 5 2 Find the value(s) of x where the tangent line is horizontal. ′ π»=π₯ π»′ = 1 πΏπ» − π»πΏ′ π′ π₯ = πΏ = 2π₯ + 5 2 πΏ′ = 2 2π₯ + 5 1 2 πΏ2 2π₯ + 5 2 (1) − (π₯)4 2π₯ + 5 πΏ′ = 4 2π₯ + 5 π′(π₯) = 2 2 2π₯ + 5 2π₯ + 5 2 − 4π₯ 2π₯ + 5 = 2π₯ + 5 4 (2π₯ + 5) 2π₯ + 5 − 4π₯ = 2π₯ + 5 4 Barnett/Ziegler/Byleen Business Calculus 12e 5 − 2π₯ = 2π₯ + 5 3 18 Continued 5 − 2π₯ π′(π₯) = 2π₯ + 5 3 The tangent line is horizontal when its slope is zero. 5 − 2π₯ 0= 2π₯ + 5 3 0 = 5 − 2π₯ 2π₯ = 5 π₯ = 2.5 The tangent line is horizontal when x=2.5 Barnett/Ziegler/Byleen Business Calculus 12e 19 Application ο§ The total cost (in hundreds of dollars) of producing x cameras per week is: πΆ π₯ = 6 + 4π₯ + 4 , 0 ≤ π₯ ≤ 30 ο§ A) Find πΆ ′ π₯ ο§ B) Find πΆ 15 πππ πΆ′(15) and interpret the results. πΆ ′ π₯ = 12 πΆ′ π₯ = 4π₯+4 −1 2 4 2 4π₯+4 −1 2 πΆ 15 = 6 + 4 β 15 + 4 = 14 πΆ ′ 15 = 2 4 β 15 + 4 −1 2 1 ′ πΆ 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
