Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 11

Answer

$F'(y)=\dfrac{14}{y^{2}}+\dfrac{9}{y^{4}}+5$

Work Step by Step

$F(y)=(\dfrac{1}{y^{2}}-\dfrac{3}{y^{4}})(y+5y^{3})$ Write the function like this for an easier differentiation process: $F(y)=(y^{-2}-3y^{-4})(y+5y^{3})$ Differentiate using the product rule $F'(y)=(y^{-2}-3y^{-4})(y+5y^{3})'+(y+5y^{3})(y^{-2}-3y^{-4})'=...$ $...=(y^{-2}-3y^{-4})(1+15y^{2})+(y+5y^{3})(-2y^{-3}+12y^{-5})=...$ Evaluate the products and simplify $y^{-2}+15-3y^{-4}-45y^{-2}-2y^{-2}+12y^{-4}-10+60y^{-2}=...$ $..=14y^{-2}+9y^{-4}+5=\dfrac{14}{y^{2}}+\dfrac{9}{y^{4}}+5$
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