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: 52


$$\frac{d}{d t} \cosh ^{-1} t=\frac{1}{\sqrt{t^{2}-1}} \quad \text { for } t>1$$

Work Step by Step

Let \begin{align*} x&=\cosh ^{-1} t \Rightarrow t=\cosh x \end{align*} then \begin{align*} \frac{d t}{d x}&=\sinh x\\ \frac{d x}{d t}&=\frac{1}{\sinh x}\\ &=\frac{1}{\sqrt{\cosh^2x-1}}\\ &=\frac{1}{\sqrt{t^2 -1}},\ \ \ t\geq 1 \end{align*} Therefore, $$\frac{d}{d t} \cosh ^{-1} t=\frac{1}{\sqrt{t^{2}-1}} \quad \text { for } t>1$$ Hence, verified.
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