Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 204: 22



Work Step by Step

$y=(x+\dfrac{1}{x})^{5}$ Let's rewrite the function like this: $y=(x+x^{-1})^{5}$ Now, differentiate using the chain rule: $y'=5(x+x^{-1})^{4}(x+x^{-1})'=5(x+\dfrac{1}{x})^{4}(1-x^{-2})$ $y'=5(x+\dfrac{1}{x})^{4}(1-\dfrac{1}{x^{2}})$
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