University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 417: 31


$- sech^{-1} x$

Work Step by Step

We are given: $y=\cos^{-1} x-x sech^{-1}x$ Recall the formula: $\dfrac{d (\cos^{-1} x)}{dx}=-\dfrac{1}{\sqrt{1- x^2}}$ and $\dfrac{d (sech^{-1} x)}{dx}=\dfrac{-1}{x \sqrt{1- x^2}}$ We need to use the product rule to get the differentiation: Thus, $\dfrac{dy}{dx}=[-\dfrac{1}{\sqrt{1- x^2}}] [ sech^{-1} x (1) +(x) \dfrac{-1}{x \sqrt{1- x^2}}]=-\dfrac{1}{\sqrt{1- x^2}}- sech^{-1} x+ \dfrac{1}{ \sqrt{1- x^2}}=- sech^{-1} x$
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