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: 36

Answer

$y'=-\sec h x\tan h x(2+\ln \sec h x)$

Work Step by Step

$\DeclareMathOperator{\sech}{sech}$ $y'=\frac{d}{dx}\sech x (1+\ln\sech x)$ Using the product rule: $y'=-\sech x\tanh x(1+\ln\sech x)+\sech x(\frac{d}{dx}(1+\ln\sech x))$ $y'=-\sech x\tanh x(1+\ln\sech x)+\sech x(\frac{d}{dx}(\ln\sech x))$ Using the chain rule: $y'=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(\frac{d\ln\sech x}{d\sech x}\times\frac{d\sech x}{dx})$ $=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(\frac{1}{\sech x}\times(-\sech x\tanh x))$ $=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(-\tanh x)$ $=-\sech x\tanh x(1+\ln\sech x)-\sech x\tanh x(1)$ $=-\sech x\tanh x(2+\ln\sech x)$
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