Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 36


$\dfrac{(\sec x \tan x)}{|\tan x|}$

Work Step by Step

Given: $y=\cosh^{-1} (\sec x)$ Since, $\dfrac{d (\cosh^{-1} x)}{dx}=\dfrac{1}{ \sqrt{x^2 -1}}$ Apply chain rule to get the differentiation. Thus, $\dfrac{dy}{d x}=\dfrac{1}{ \sqrt{sec^2 x-1}}(\sec x \tan x)=\dfrac{1}{ \sqrt{ tan^2 x}}(\sec x \tan x)$ or, $\dfrac{dy}{d x}=\dfrac{(\sec x \tan x)}{|\tan x|}$
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