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: 48


$$\frac{d}{d x} \operatorname{sech} x=-\operatorname{sech} x \tanh x$$

Work Step by Step

Since $$ \operatorname{sech} x =\frac{2}{e^x+e^{-x}}$$ Then \begin{align*} \frac{d}{d x} \operatorname{sech} x&= \frac{d}{d x}\frac{2}{e^x+e^{-x}}\\ &= 2\frac{d}{dx}\left(\left(e^x+e^{-x}\right)^{-1}\right)\\ &=-\frac{2\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}}\\ &= -\frac{ \left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right) } \frac{2}{\left(e^{x}+e^{-x}\right) }\\ &= -\operatorname{sech} x \tanh x \end{align*}
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