Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 40

Answer

$$y'=|\sec x|$$

Work Step by Step

$y'=\frac{d}{dx}\sinh^{-1}(\tan x)$ Using the chain rule: $y'=\frac{d\sinh^{-1}(\tan x)}{d\tan x} \times \frac{d\tan x}{dx}$ $=\frac{1}{\sqrt{1+(\tan x)^2}} \times \sec^2 x$ Recall that: $$\tan^2 x+1=\sec^2 x$$ Thus, $y'=\frac{1}{\sqrt{1+\tan^2 x}} \times \sec^2 x$ $=\frac{1}{|\sec x|}\times \sec^2 x$ $=|\sec x|$
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