<s>[INST] Two differentiation results of $\sin^{-1}(2x\sqrt{1-x^2})$ While trying to differentiate $\sin^{-1}(2x\sqrt{1-x^2})$, if we put
$x = \sin\theta$, we get,
\begin{align*}
  y &=\sin^{-1}(2x\sqrt{1-x^2})\\
  &= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
  &= \sin^{-1}(2\sin\theta\cos\theta)\\
  &= \sin^{-1}(\sin2\theta)\\
  &= 2\theta\\
  &= 2\sin^{-1}x.
\end{align*}
So,
\begin{align*}
  \frac{dy}{dx} &= \frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
But if we put $x = \cos\theta$, we get,
\begin{align*}
  y &=\sin^{-1}(2x\sqrt{1-x^2})\\
  &= \sin^{-1}(2\cos\theta\sqrt{1-\cos^2\theta})\\
  &= \sin^{-1}(2\cos\theta\sin\theta)\\
  &= \sin^{-1}(\sin2\theta)\\
  &= 2\theta\\
  &= 2\cos^{-1}x.
\end{align*}
This time, 
\begin{align*}
  \frac{dy}{dx} &= -\frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
We are perplexed about the difference in sign between the two results
and thought that you could help. 
(We understand that we can differentiate $\sin^{-1}(2x\sqrt{1-x^2})$ directly, without any substitution, which gives us the first result.)
 [/INST] This Question bugged me for long time before I found the linked identity
Setting  $y=x$ in Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x\\
\begin{align}
&=\arcsin( 2x\sqrt{1-x^2}) \;\;;2x^2 \le 1  \\
&=\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1\text{ and } 0< x\\
&=-\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1\text{ and }  x\le 0\\
\end{align}
$$
Case $\#1:$ If $2x^2 \le 1,\arcsin( 2x\sqrt{1-x^2})=2\arcsin x$
Case $\#2:$  If $2x^2> 1,$
Case $\#2A: x>0,\arcsin( 2x\sqrt{1-x^2})=\pi-2\arcsin x=2\arccos x$ 
Case $\#2B: x\le0,\arcsin( 2x\sqrt{1-x^2})=-\pi-2\arcsin x$ 
</s>