Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Review - Exercises - Page 270: 51

Answer

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

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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