Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 28

Answer

$J'(v)=1+u^{-2}+6u^{-4}$

Work Step by Step

$J(v)=(v^3-2v)(v^{-4}+v^{-2})$ Use the product rule. $J'(v)=(v^3-2v)(-4v^{-5}-2v^{-3})+(v^{-4}+v^{-2})(3v^2-2)$ Distribute. $J'(v)=-4v^{-2}-2+8v^{-4}+4v^{-2}+3v^{-2}-2v^{-4}+3-2v^{-2}$ Combine like terms. $J'(v)=1+u^{-2}+6u^{-4}$
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