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: 47


See the proof below.

Work Step by Step

Using the definition of $\tanh x$ and the quotient rule, we have \begin{align*} \frac{d}{d x} \tanh x &=\frac{d}{d x}\left(\frac{\sinh x}{\cosh x}\right) \\ &=\frac{\cosh x \cdot \sinh x^{\prime}-\sinh x \cdot \cosh ^{\prime} x}{(\cosh x)^{2}} \\ &=\frac{\cosh x \cdot \cosh x-\sinh x \cdot \sinh x}{\cosh ^{2} x} \\ &=\frac{\cosh ^{2} x-\sinh ^{2} x}{\cosh ^{2} x} \\ &=1-\tanh ^{2} x \\ &=\operatorname{sech}^{2} x \end{align*} where we used the fact that $\left(\tanh ^{2} x+\operatorname{sech}^{2} x=1\right) .$
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