## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 14

#### Answer

$$y'=\cosh x\tanh x+\sinh x\, sech^2 x .$$

#### Work Step by Step

Since $y=\sinh x\tanh x$, then the derivative $y'$, using the product rule $(uv)'=uv'+u'v$ and the facts that $(\sinh x)'=\cosh x$ and $(\tanh x)'=sech^2x$, is given by $$y'=\cosh x\tanh x+\sinh x\, sech^2 x .$$

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