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: 17


$$ y'= \frac{1}{x}\cosh (\ln x).$$

Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\cosh x)'=\sinh x$ Since $ y=\sinh (\ln x) $, then the derivative, by using the chain rule, is given by $$ y'=\cosh (\ln x) (\ln x)'= \frac{1}{x}\cosh (\ln x).$$
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