## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 22

#### Answer

$\text{coth}^3 v$

#### Work Step by Step

Use formulas: Since, $\dfrac{d}{dx} (\sinh x)=\cosh x$; $\dfrac{d}{dx} (coth x)=-csch^2 x$ and $coth^2 x -csch^2 x=1$ $\dfrac{dy}{dv} =\dfrac{d}{dv} [\ln \sinh v-\dfrac{1}{2} \text{coth}^2 v)]=\dfrac{1}{\sinh v}(\cosh v)-\dfrac{1}{2}(2) \text{coth} v (csch^2 v)$ Thus, $\dfrac{dy}{dv}=\text{coth} v (1+csch^2 v)=\text{coth}^3 v$

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