## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 265: 45

#### Answer

$$y'=-\csc x$$

#### Work Step by Step

$y'=\frac{d}{dx}\coth^{-1}(\sec x)$ Using the chain rule: $y'=\frac{d\coth^{-1}(\sec x)}{d\sec x} \times \frac{d\sec x}{dx}$ $=\frac{1}{1-\sec^2 x}\times\sec x\tan x$ Recall that: $$\sec^2 x-1=\tan^2 x$$ Thus, $y'=\frac{1}{-\tan^2 x}\times\sec x\tan x$ $=\frac{-\sec x}{\tan x}$ $=\frac{\frac{-1}{\cos x}}{\frac{\sin x}{\cos x}}$ $=\frac{-1}{\sin x}$ $=-\csc x$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.