Differentiate $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$ 
Find $\frac{dy}{dx}$ if $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$

The solution is given as $y'=0$ in my reference. But that doesn't seem to be a complete solution as the graph of the function is:

My Attempt
Let $x=\sin\alpha\implies \alpha=\sin^{-1}x$
$$
y=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{1-\sin^2\alpha}=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\cos^2\alpha}\\
=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\sin^2(\tfrac{\pi}{2}-\alpha)}=\sin^{-1}(\sin\alpha)+\sin^{-1}|\sin(\tfrac{\pi}{2}-\alpha)|\\
=n\pi+(-1)^n(\alpha)+
$$
How do I proceed further and find the derivative ?
 A: The derivative is: (for positive x)
$$y^{\prime}=\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{\frac{-2x}{2\sqrt{1-x^{2}}}}{\sqrt{1-1+x^{2}}}=\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{-1}{\sqrt{1-x^{2}}}=0$$
if x<0 we have :
$$y^{\prime}=\dfrac{2}{\sqrt{1-x^{2}}}$$
A: From the end of your second line you obtain
$$
\sin^{-1}(\sin\alpha)+\sin^{-1}\left(\sin\left(\frac\pi2-\alpha\right)\right)=\alpha+\frac\pi2-\alpha=\frac\pi2
$$
for $\alpha\in[0,\pi/2]$ and
$$
\alpha+\alpha-\frac\pi2=2\sin^{-1}(x)-\frac\pi2
$$
for $\alpha\in[-\pi/2,0]$, which gives the same derivatives as in a standard method.
A: $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
Let, $x=\sin\alpha\implies \alpha=\sin^{-1}x$, We have $-\pi/2\leq\alpha\leq\pi/2\implies|\cos\alpha|=\cos\alpha$
$$
\begin{align}
y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(|\cos\alpha|)=\sin^{-1}(\sin\alpha)+\sin^{-1}(\cos\alpha)\\&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))
\end{align}
$$
Here,
$$
\tfrac{-\pi}{2}\leq\alpha\leq\tfrac{\pi}{2}\implies\sin^{-1}(\sin\alpha)=\alpha\\
0\leq\tfrac{\pi}{2}-\alpha\leq{\pi}\implies\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))=\begin{cases}\frac{\pi}{2}-\alpha,\text{ if }0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\
\pi-(\frac{\pi}{2}-\alpha),\text{ if }\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi\end{cases}
$$
Therefore,
$$
\begin{align}
y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))\\
&=\begin{cases}\alpha+\frac{\pi}{2}-\alpha=\frac{\pi}{2}\quad\quad\quad\quad\;\;\text{ if }\quad\: 0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\
\alpha+\pi-\frac{\pi}{2}+\alpha=\frac{\pi}{2}+2\alpha\quad\text{ if }\quad\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi
\end{cases}\\
&=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\text{ if }\quad0\leq\alpha\leq\tfrac{\pi}{2}\\
\frac{\pi}{2}+2\alpha\quad\text{ if }\quad \frac{-\pi}{2}\leq\alpha<0
\end{cases}\\
&=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\quad\quad\text{ if }\quad\quad 0\leq x\leq 1\\
\frac{\pi}{2}+2\sin^{-1}x\quad\text{ if }\quad -1\leq x<0
\end{cases}
\end{align}
$$
$$
\color{red}{
\frac{dy}{dx}=\begin{cases}0\quad\quad\quad\text{ if }\;\quad 0\leq x\leq 1\\
\frac{2}{\sqrt{1-x^2}}\quad\;\text{ if }\; -1\leq x<0
\end{cases}}
$$
