Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 40


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

Work Step by Step

Given $$y=\tan^{-1}\left(\sin^{-1}\sqrt{x}\right)$$ Then \begin{align*} y'&=\frac{1}{1+(\sin^{-1}\sqrt{x})^2}\frac{1}{\sqrt{1-x}}\frac{1}{2\sqrt{x}}\\ &=\frac{1}{2\sqrt{x}\left((\sin^{-1}\sqrt{x})^2+1\right)\sqrt{1-x}} \end{align*}
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