Calculus 8th Edition

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

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 32


$$y' =\frac{4\sin ^{-1}2x}{\sqrt{1-4x^2}}$$

Work Step by Step

Given $$y=\left(\sin ^{-1}2x\right)^2$$ Then \begin{align*} y'&=2\left(\sin ^{-1}2x\right) \frac{2}{\sqrt{1-4x^2}}\\ &=\frac{4\sin ^{-1}2x}{\sqrt{1-4x^2}} \end{align*}
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