Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 44

Answer

$$y'=\frac{1}{\sqrt{1-e^{-2x}}}$$

Work Step by Step

$\DeclareMathOperator{\sech}{sech}$ $y'=\frac{d}{dx}\sech^{-1}(e^{-x})$ Using the chain rule: $y'=\frac{d\sech^{-1}(e^{-x})}{de^{-x}} \times\frac{de^{-x}}{dx}$ $=-\frac{1}{{e^{-x}}\sqrt{1-({e^{-x}})^2}} \times-e^{-x}$ $=\frac{e^{-x}}{e^{-x}\sqrt{1-e^{-2x}}}$ $=\frac{1}{\sqrt{1-e^{-2x}}}$
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