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: 21


$$ y'= -\frac{1}{2\sqrt{x}} sech(\sqrt{x}) \tanh (\sqrt{x}) .$$

Work Step by Step

Since $ y=sech(\sqrt{x})$, then the derivative, by using the chain rule and the facts that $(\ sech \ x)'= -\ sech \ x \tanh x $ and $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$, is given by $$ y'=-sech(\sqrt{x}) \tanh (\sqrt{x}) (\sqrt{x})'=- \frac{1}{2\sqrt{x}} sech(\sqrt{x}) \tanh (\sqrt{x}) .$$
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