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: 27


$-\tanh^{-1} \theta +\dfrac{1}{(1 + \theta)}$

Work Step by Step

Given: $y=(1-\theta) \tanh^{-1} \theta$ Since, $\dfrac{d (\tanh^{-1} x)}{dx}=\dfrac{1}{1-x^2}$ Apply product rule to get the differentiation. $\dfrac{dy}{d \theta}=\tanh^{-1} \theta (-1)+(1-\theta) \dfrac{1}{1- \theta^2}=-\tanh^{-1} \theta+(1-\theta) [\dfrac{1}{(1- \theta)(1 + \theta)}]$ or, $=-\tanh^{-1} \theta +\dfrac{1}{(1 + \theta)}$
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