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. β’ π π₯ = 3π₯ + 4 β’ π π₯ = π₯2 β’ π π₯ = π π(π₯) ο§ Ex 2: π π₯ = π 4π₯β1 ο§ We can express π π₯ as a composition of functions. β’ π π₯ = 4π₯ β 1 β’ π π₯ = ππ₯ β’ 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: β² β’ π¦=π π₯ ο y = ππ¦ ππ₯ ο§ u is a function of x: = ππ’ ππ₯ β’ π¦ = π(π’) ο y = ππ¦ ππ’ β’ π’ = π(π₯) ο uβ² ο§ y is a function of u: β² 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 βoutside-insideβ 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¦ β² = πβ²(π π₯ ) β πβ²(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 β² =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¦β² 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₯ β 2π₯ 7 ππ’ π₯ π’ = π β 2π₯ = ππ₯ β 2 ππ₯ ππ¦ 7 π¦=π’ = 7π’6 ππ’ ππ¦ ππ’ π¦β² = β ππ’ ππ₯ ππ¦ ππ’ π¦β² = β ππ’ ππ₯ = 7 π π₯ β 2π₯ Ex 7: π¦ = 5π β4π₯ ππ’ π’ = β4π₯ = β4 ππ₯ ππ¦ π’ = 5π π’ π¦ = 5π ππ’ 6 β ππ₯ β 2 Barnett/Ziegler/Byleen Business Calculus 12e = 5π β4π₯ (β4) = β20π β4π₯ 9 Derivatives Using the Chain Rule Ex. 8: π¦ = π’= 2π₯ 3 π¦=8 π¦β² 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’ β² 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 β² π¦ = 2 β 6π₯ 3π₯ 6π₯ β² π¦ = 2 3π₯ πΈπ₯. 11 π¦ = ln π₯ 2 + 2π₯ + 1 2 β² π¦ = π₯ 1 π¦β² = 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₯ β7 π¦ = π₯2 β 7 4 β4 π¦β² = β4 π₯ 2 β 7 β5 = β8π₯ π₯ 2 β 7 Barnett/Ziegler/Byleen Business Calculus 12e 2π₯ β5 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₯ β 4 5 π β² π₯ = πΏπ β² + π πΏβ² L Rβ² R π β² π₯ = π₯ 2 β 5 3π₯ β 4 4 β 3 + 3π₯ β 4 π β² π₯ = 15π₯ 2 3π₯ β 4 4 + 2π₯ 3π₯ β 4 Lβ² 5 β 2π₯ 5 (ππππ‘ππππ ππ’π‘ ππ πππ‘β π‘ππππ ) π β² π₯ = π₯ 3π₯ β 4 4 15π₯ + 2(3π₯ β 4) π β² π₯ = π₯ 3π₯ β 4 4 15π₯ + 6π₯ β 8 π β² π₯ = π₯ 3π₯ β 4 4 21π₯ β 8 Barnett/Ziegler/Byleen Business Calculus 12e 15 Quotient Rule w/ Chain Rule π₯4 π π₯ = 3π₯ β 8 β² 2 π» = π₯4 πΏ = 3π₯ β 8 πΏπ» β π»πΏβ² π₯ = πΏ2 3π₯ β 8 2 4π₯ 3 β π₯ 4 6 3π₯ β 8 πβ²(π₯) = 2 2 3π₯ β 8 πβ² 4π₯ 3 3π₯ β 8 2 β 6π₯ 4 3π₯ β 8 = 3π₯ β 8 4 2π₯ 3 (3π₯ β 8) 2 3π₯ β 8 β 3π₯ = 3π₯ β 8 4 Barnett/Ziegler/Byleen Business Calculus 12e π»β² = 4π₯ 3 2 πΏβ² = 2 3π₯ β 8 1 3 πΏβ² = 6 3π₯ β 8 2π₯ 3 3π₯ β 16 = 3π₯ β 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 πβ² π₯ = 2π₯ + 8 β1 2 2 2 = 2π₯ + 8 β1 2 π 4 =4 πβ² 1 4 = 4 Barnett/Ziegler/Byleen Business Calculus 12e 4,4 π = 14 π¦ β 4 = 14 π₯β4 π¦ β 4 = 14π₯β1 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. β² π»=π₯ π»β² = 1 πΏπ» β π»πΏβ² πβ² π₯ = πΏ = 2π₯ + 5 2 πΏβ² = 2 2π₯ + 5 1 2 πΏ2 2π₯ + 5 2 (1) β (π₯)4 2π₯ + 5 πΏβ² = 4 2π₯ + 5 πβ²(π₯) = 2 2 2π₯ + 5 2π₯ + 5 2 β 4π₯ 2π₯ + 5 = 2π₯ + 5 4 (2π₯ + 5) 2π₯ + 5 β 4π₯ = 2π₯ + 5 4 Barnett/Ziegler/Byleen Business Calculus 12e 5 β 2π₯ = 2π₯ + 5 3 18 Continued 5 β 2π₯ πβ²(π₯) = 2π₯ + 5 3 The tangent line is horizontal when its slope is zero. 5 β 2π₯ 0= 2π₯ + 5 3 0 = 5 β 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 β€ π₯ β€ 30 ο§ A) Find πΆ β² π₯ ο§ B) Find πΆ 15 πππ πΆβ²(15) and interpret the results. πΆ β² π₯ = 12 πΆβ² π₯ = 4π₯+4 β1 2 4 2 4π₯+4 β1 2 πΆ 15 = 6 + 4 β 15 + 4 = 14 πΆ β² 15 = 2 4 β 15 + 4 β1 2 1 β² 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