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: 32

Answer

$- sech^{-1}x \dfrac{x}{\sqrt{1- x^2}}$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.