Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 24


$$ y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}$$

Work Step by Step

Recall that $(\sinh x)'=\cosh x$ Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\ln x)'=\dfrac{1}{x}$ Since $ y= x^{\sinh x}$, then taking the $\ln $ on both sides, we get $$\ln y=\sinh x \ln x.$$ Now, the derivative, by using the product rule, is given by $$ y'/y= \cosh x \ln x+ \frac{\sinh x}{x}\Longrightarrow y'= y(\cosh x \ln x+ \frac{\sinh x}{x}) $$ and hence $$ y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.