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: 26


$\dfrac{1}{ \sqrt{(x+1)(4x+3)}} $

Work Step by Step

Given: $y=\cosh^{-1} 2\sqrt {x+1}$ Since, $\dfrac{d (\cosh^{-1} x)}{dx}=\dfrac{1}{\sqrt{x^2-1}}$ Then, $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{(2\sqrt {x+1}-1)^2}} (\dfrac{2}{2 \sqrt {x+1}})=\dfrac{1}{\sqrt{4(x+1)-1}} (\dfrac{1}{ \sqrt {x+1}})$ or, $=\dfrac{1}{ \sqrt{(x+1)(4x+3)}} $
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