Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get: 
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)^n(a-b)=n\pi+(-1)^n(\sin^{-1}x-\sin^{-1}\sqrt{x})
$$
Thus,
$$
y'=\frac{d}{dx}\big[n\pi+(-1)^n(a-b)\big]=\begin{cases}\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\text{ if }n\text{ is even}\\
-\bigg[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\bigg]\text{ if }n\text{ is odd}
\end{cases}
$$
Is it the right way to solve this problem and how do I check the solution is correct ?
Note: I think there got to be two cases for the derivative as the graph of the function is

 A: First let $g(x) = x\sqrt{1-x} + \sqrt{x}\sqrt{1-x^2}$. Then $f(x) = \sin^{-1}(g(x))$. Now we use the chain rule, so $f'(x) = \frac{1}{\sqrt{1-g(x)^2}}g'(x)$. I let you finish, by finding $g'$.
A: Use that $$(\arcsin(x))'=\frac{1}{\sqrt{1-x^2}}$$
the whole derivative is given by $$f'(x)={\frac {1}{\sqrt {- \left( x\sqrt {1-x}+\sqrt {x}\sqrt {-{x}^{2}+1}
 \right) ^{2}+1}} \left( \sqrt {1-x}-1/2\,{\frac {x}{\sqrt {1-x}}}+1/2
\,{\frac {\sqrt {-{x}^{2}+1}}{\sqrt {x}}}-{\frac {{x}^{3/2}}{\sqrt {-{
x}^{2}+1}}} \right) }
$$
after simplifying i got:
$$\frac {\sqrt {x} \left (3 x
               \left (\sqrt {x} + \sqrt {x + 1} \right) - 2
               \sqrt {x + 1} \right) - 1} {2 \sqrt {(-x + 1) x}
    \sqrt {x + 1} \sqrt {2 (x - 1) \sqrt {x + 1} x^{3/2} + 2
        x^3 - x^2 - x + 1}}$$
A: $$F(x)=\sin^{-1}x-\sin^{-1}\sqrt x=\sin^{-1}x+\sin^{-1}(-\sqrt x)$$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$\displaystyle F(x)  =\begin{cases}
\arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x \le 1 \;\text{ or }\; x^2+x > 1, -x\sqrt x< 0\iff x>0\\
\pi - \arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x > 1, 0< x,-\sqrt x \le 1\text{ which is untenable}\\
-\pi - \arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x > 1, -1< x,-\sqrt x \le 0
\end{cases}$$
A: Let $\sin a=x\implies a=\sin^{-1}x$ and $\sin b=\sqrt{x}\implies b=\sin^{-1}\sqrt{x}$
$$
y=\sin^{-1}\Big[ \sin a|\cos b|-\sin b.|\cos a| \Big]\\
$$
$0<x<1\implies 0<\sin^{-1}x=a<\frac{\pi}{2}\implies |\cos a|=\cos a$ and
$0<x<1\implies 0<\sqrt{x}<1\implies0<\sin^{-1}\sqrt{x}=b<\frac{\pi}{2}\implies|\cos b|=\cos b$
$$
\begin{align}
y&=\sin^{-1}\Big[\sin a\cos b-\cos a\sin b\Big]\\
&=\sin^{-1}\big[\sin(a-b)\big]
\end{align}
$$
We have $0<x<\frac{\pi}{2}$ and $0<b<\frac{\pi}{2}\implies \frac{-\pi}{2}<-b<0$, Hence $\frac{-\pi}{2}<a-b<\frac{\pi}{2}$
$$
y=\sin^{-1}\big[\sin(a-b)\big]=a-b=\sin^{-1}x-\sin^{-1}\sqrt{x}
$$
$$
\begin{align}
\color{blue}{\frac{dy}{dx}}&\color{blue}{=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}}\\
&=\frac{2\sqrt{x}\sqrt{1-x}-\sqrt{1-x^2}}{2\sqrt{x}\sqrt{1-x}\sqrt{1-x^2}}
=\frac{\sqrt{1-x}(2\sqrt{x}-\sqrt{1+x})}{2\sqrt{x}\sqrt{1-x}\sqrt{1-x^2}}
\end{align}
$$
$y'<0$ when $2\sqrt{x}<\sqrt{1+x}\implies 4x<1+x\implies 0<x<\frac{1}{3}$
$y'<0$ when $\frac{1}{3}<x<1$
