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*}
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