Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 265: 41



Work Step by Step

$y'=\frac{d}{dx}\cosh^{-1}{\sqrt{x}}$ Using the chain rule: $y'=\frac{d\cosh^{-1}{\sqrt{x}}}{d\sqrt{x}} \times \frac{d\sqrt{x}}{dx}$ $=\frac{1}{\sqrt{(\sqrt x)^2-1}} \times \frac{1}{2\sqrt x}$ $=\frac{1}{2\sqrt x\sqrt{x-1}}$ $=\frac{1}{2\sqrt{x^2-x}}$
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