Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 160: 17


$g'(y) = 18y^5-52y^3+8y$

Work Step by Step

$g(y) = (3y^4-y^2)(y^2-4)$ a) Using Product Rule: $g'(y) = (12y^3 -2y)(y^2-4)+(3y^4-y^2)(2y) = 12y^5-2y^3-48y^3+8y +6y^5 - 2y^3 = 18y^5-52y^3+8y$ b) Expansion then Power Rule: $g(y) = 3y^6-12y^4-y^4+4y^2 = 3y^6 - 13y^4 + 4y^2$ $g'(y) = 18y^5-52y^3+8y$ Answers to Part A and Part B match
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