Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 46


$$y'=- \frac{1+\sqrt{1-x^2}}{ \sqrt{1-x^2}\sqrt{1-(x+\sin^{-1}x)^2}}.$$

Work Step by Step

Since $ y=\cos^{-1}(x+\sin^{-1}x) $, then $$ y'=- \frac{1}{ \sqrt{1-(x+\sin^{-1}x)^2}}(x+\sin^{-1}x)'\\ =- \frac{1+1/\sqrt{1-x^2}}{ \sqrt{1-(x+\sin^{-1}x)^2}}\\ =- \frac{1+\sqrt{1-x^2}}{ \sqrt{1-x^2}\sqrt{1-(x+\sin^{-1}x)^2}}.$$
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