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: 34


$-\dfrac{\ln 2}{ \sqrt{1+4^\theta}}$

Work Step by Step

Given: $y=csch^{-1} 2^{\theta}$ Since, $\dfrac{d (csch^{-1} x)}{dx}=\dfrac{-1}{|x| \sqrt{1 + x^2}}$ and $\dfrac{d m^x}{dx}=m^x \ln m$ Apply product rule to get the differentiation. Thus, $\dfrac{dy}{d \theta}=\dfrac{-1}{|2^{\theta}| \sqrt{1 + ((2)^{\theta})^2}}(2)^{\theta} (\ln 2)$ or, $\dfrac{dy}{d \theta}=-\dfrac{\ln 2}{ \sqrt{1+4^\theta}}$
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