University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 417: 22


$\text{coth}^3 v$

Work Step by Step

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