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Chapter 4 Additional Derivative Topics Section 3 Derivatives of Products and Quotients Objectives for Section 4.3 Derivatives of Products and Quotients The student will be able to calculate: β the derivative of a product of two functions, and β the derivative of a quotient of two functions. Barnett/Ziegler/Byleen Business Calculus 12e 2 Review Exponential & Log Derivatives Find the derivative of each function: π¦ = ππ₯ π¦β² = π π₯ π¦ = ln π₯ 1 π¦β² = π₯ π¦ = ππ₯ π¦β² = π π₯ ln π π¦ = log π π₯ 1 π¦β² = π₯ ln π Barnett/Ziegler/Byleen Business Calculus 12e 3 Example 1 Derivative of a Product Sometimes the derivative of a product can be found by first using the distributive property. Find the derivative of y = 5x2(x3 + 2). π¦ = 5π₯ 5 + 10π₯ 2 π¦ β² = 25π₯ 4 + 20π₯ But what about a product like this one? π¦ = 5π₯ 8 π π₯ You MUST use the Product Ruleβ¦ Barnett/Ziegler/Byleen Business Calculus 12e 4 Derivatives of Products Theorem 1 (Product Rule) If f (x) = L(x) ο R(x), and if Lο’(x) and Rο’(x) exist, then f ο’(x) = L(x) ο Rο’(x) + R(x) ο Lο’(x) Mnemonic: Left D Right + Right D Left πΏπ β² + π πΏβ² Barnett/Ziegler/Byleen Business Calculus 12e 5 Example 2 ο§ π π₯ = 5π₯ 8 π π₯ Find πβ²(π₯) πΏ π₯ = 5π₯ 8 π π₯ = ππ₯ πΏβ² π₯ = 40π₯ 7 π β² π₯ = π π₯ π β² (π₯) = πΏ π₯ β π β² π₯ + π (π₯) β πΏβ²(π₯) π β² (π₯) = 5π₯ 8 β π π₯ + π π₯ β 40π₯ 7 π β² (π₯) = 5π₯ 7 π π₯ (π₯ + 8) Barnett/Ziegler/Byleen Business Calculus 12e 6 Example 3 ο§ π¦ = π₯ + 2 3π₯ Find yβ² πΏ π₯ =π₯+2 π π₯ = 3π₯ πΏβ² π₯ = 1 π β² π₯ = 3π₯ ln 3 π¦ β² = πΏ π₯ β π β² π₯ + π (π₯) β πΏβ²(π₯) π¦ β² = π₯ + 2 (3π₯ ln 3) + 3π₯ β 1 π¦ β² = π₯ + 2 (3π₯ ln 3) + 3π₯ π¦ β² = 3π₯ [ π₯ + 2 ln 3 + 1] Barnett/Ziegler/Byleen Business Calculus 12e 7 Derivatives of Quotients ο§ Some derivatives of quotients can be found by rewriting the function: 4π₯ 3 β 8π₯ 2 π β² π₯ = 4π₯ β 4 π π₯ = 2π₯ = 2π₯ 2 β 4π₯ ο§ You canβt do that with functions like this: π‘ 3 β 3π‘ π¦= 2 π‘ β4 Barnett/Ziegler/Byleen Business Calculus 12e 8 Derivatives of Quotients Theorem 2 (Quotient Rule) π»(π₯) If f (x) = , and if Hο’(x) and Lο’(x) exist, then πΏ(π₯) β² π₯ β π»(π₯) β πΏβ²(π₯) πΏ π₯ β π» πβ² π₯ = πΏ(π₯)2 Mnemonic: πΏππ€ π· π»ππββπ»ππβ π· πΏππ€ πΏππ€ 2 πΏπ» β² β π»πΏβ² πΏ2 Barnett/Ziegler/Byleen Business Calculus 12e 9 Example 4 π‘ 3 β 3π‘ Find the derivative of: π¦ = 2 π‘ β4 π» π‘ = π‘ 3 β 3π‘ πΏ π‘ = π‘2 β 4 π» β² π‘ = 3π‘ 2 β 3 πΏβ² π‘ = 2π‘ β² π‘ β π»(π‘) β πΏβ²(π‘) πΏ π‘ β π» π¦β² = πΏ(π‘)2 π¦β² π‘ 2 β 4 3π‘ 2 β 3 β π‘ 3 β 3π‘ 2π‘ = π‘2 β 4 2 4 β 3π‘ 2 β 12π‘ 2 + 12 β 2π‘ 4 β 6π‘ 2 3π‘ π¦β² = π‘2 β 4 2 Barnett/Ziegler/Byleen Business Calculus 12e 4 β 9π‘ 2 + 12 π‘ π¦β² = π‘2 β 4 2 10 Example 5 Find the derivative of π¦ = π» π₯ = ln π₯ 1 β² π» π₯ = π₯ πΏ π₯ β πΏ π₯ = 2π₯ + 5 πΏβ² π₯ = 2 π»β² π₯ β π»(π₯) β πΏβ²(π₯) = πΏ(π₯)2 1 2π₯ + 5 β β (ln π₯) β 2 π₯ π¦β² = 2π₯ + 5 2 π¦β² ln π₯ 2π₯+5 2π₯ + 5 β 2π₯ ln π₯ π₯ π¦β² = 2π₯ + 5 2 2π₯ + 5 β 2π₯ln π₯ π¦ = π₯ 2π₯ + 5 2 β² 2π₯ + 5 β 2 ln π₯ π₯ β² π¦ = 2π₯ + 5 2 Barnett/Ziegler/Byleen Business Calculus 12e 11 Homework Barnett/Ziegler/Byleen Business Calculus 12e 12 Objectives for Section 4.3 Day 2 The student will be able to : β Review derivatives of products and quotients πΏπ»β² β π»πΏβ² π¦β² = πΏ2 β Find the equations of tangent lines β Solve applications Barnett/Ziegler/Byleen Business Calculus 12e 13 Review ο§ What is the mnemonic for finding the derivative of a product π¦ = πΏπ ? ο§ π¦ β² = πΏπ β² + π πΏβ² ο§ What is the mnemonic for finding the derivative of a π» quotient π¦ = ? πΏ β² ο§ π¦ = πΏπ» β² βπ»πΏβ² πΏ2 Barnett/Ziegler/Byleen Business Calculus 12e 14 Example 1 ο§ Find ββ² π₯ , where f(x) is an unspecified differentiable function. β π₯ = 2π₯ 4 π(π₯) ββ² π₯ = πΏπ β² + π πΏβ² ββ² π₯ = 2π₯ 4 β π β² π₯ + π(π₯) β 8π₯ 3 ββ² π₯ = 2π₯ 3 π₯π β² π₯ + 4π(π₯) Barnett/Ziegler/Byleen Business Calculus 12e Factor 2π₯ 3 from each term. 15 Example 2 ο§ Find ββ² π₯ , where f(x) is an unspecified differentiable function. β² πΏπ» β π»πΏβ² β² π(π₯) β π₯ = 2 β π₯ = 3 πΏ π₯ 3 πβ²(π₯) β π(π₯)3π₯ 2 π₯ ββ² π₯ = π₯3 2 Divide each term by π₯ 2 . 3 πβ²(π₯) β π(π₯)3π₯ 2 π₯ ββ² π₯ = π₯6 ββ² π₯πβ²(π₯) β 3π π₯ π₯ = π₯4 Barnett/Ziegler/Byleen Business Calculus 12e 16 Example 3 Let f (x) = (x2 + 6)(ln x). Find the equation of the line tangent to the graph of f (x) at x = 1. 1 π π₯ = π₯ + 6 + ln π₯ 2π₯ π₯ 1 β² π 1 = 1 + 6 + ln 1 2(1) 1 πβ² 1 = 7 + 0 β 2 = 7 π¦ β π¦1 = π(π₯ β π₯1 ) π 1 = (7)(ln 1) = 7 β 0 = 0 π¦ β 0 = 7(π₯ β 1) β² 2 π=7 (1,0) π¦ = 7π₯ β 7 Barnett/Ziegler/Byleen Business Calculus 12e 17 Example 4 3π₯ . 2π₯β1 Let f (x) = Find the equation of the line tangent to the graph of f (x) at x = 2. πβ² 2π₯ β 1 3 β 3π₯(2) π₯ = (2π₯ β 1)2 πβ² β3 6π₯ β 3 β 6π₯ = π₯ = (2π₯ β 1)2 (2π₯ β 1)2 β3 β1 π 2 = = 9 3 β² 2, π 2 = (2,2) π¦ β π¦1 = π(π₯ β π₯1 ) β1 π¦β2= (π₯ β 2) 3 β1 8 π¦= π₯+ 3 3 Barnett/Ziegler/Byleen Business Calculus 12e 18 Application 1 ο§ The total sales (in thousands of games) of Call of Duty x months after the game is introduced is given by: 150π₯ π π₯ = π₯+3 ο§ A) Find π β² π₯ ο§ B) Find π 12 and π β² 12 and interpret the results. ο§ C) Use the results from (B) to estimate the total sales after 13 months. Barnett/Ziegler/Byleen Business Calculus 12e 19 150π₯ π π₯ = π₯+3 Application 1 (continued) ο§ A) Find π β² π₯ β² β π»πΏβ² πΏπ» πβ² π₯ = πΏ2 πβ² π₯ + 3 150 β 150π₯(1) π₯ = (π₯ + 3)2 πβ² 150π₯ + 450 β 150π₯ π₯ = (π₯ + 3)2 πβ² 450 π₯ = (π₯ + 3)2 Barnett/Ziegler/Byleen Business Calculus 12e 20 Application 1 (continued) ο§ B) Find π 12 and π β² 12 and interpret the results 150π₯ π π₯ = π₯+3 π 12 = 150(12) 12 + 3 πβ² 450 π₯ = (π₯ + 3)2 π β² 12 = 450 (12 + 3)2 1800 π 12 = 15 450 π 12 = 225 π 12 = 120 π β² 12 = 2 β² After 12 months, the total sales are 120,000 games and sales are increasing at a rate of 2,000 games per month. Barnett/Ziegler/Byleen Business Calculus 12e 21 Application 1 (continued) ο§ C) Use the results from (B) to estimate the total sales after 13 months. π 12 + πβ²(12) 120 + 2 = 122 After 13 months, the total sales will be 122,000 games. Barnett/Ziegler/Byleen Business Calculus 12e 22 Homework #4-3B: Pg 231 31, 35, 39, 43, 49, 51, 81, 83, 87 Barnett/Ziegler/Byleen Business Calculus 12e 23