Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 44



Work Step by Step

Given: $\DeclareMathOperator{\sech}{sech}$$y=\sech^{-1}(e^{-x})$ Now, $y'=\dfrac{d}{dx}\sech^{-1}(e^{-x})$ Apply chain rule, we have $y'=-\dfrac{1}{{e^{-x}}\sqrt{1-({e^{-x}})^2}}(-e^{-x})$ or, $y'=\dfrac{e^{-x}}{e^{-x}\sqrt{1-e^{-2x}}}$ Thus, $y'=\dfrac{1}{\sqrt{1-e^{-2x}}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.