Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 267: 47


$y'=\dfrac{\cosh x}{\sqrt{\sinh^{2}x-1}}$

Work Step by Step

$y=\cosh^{-1}(\sinh x)$ Start the differentiation process by using the chain rule: $y'=\dfrac{1}{\sqrt{\sinh^{2}x-1}}(\sinh x)'=...$ Evaluate the remaining derivative and simplify: $...=\dfrac{\cosh x}{\sqrt{\sinh^{2}x-1}}$
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