Answer
$$y' =\frac{4\sin ^{-1}2x}{\sqrt{1-4x^2}}$$
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*}
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 32
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