<s>[INST] 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

 [/INST] 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:
$$\frac {\sqrt {x} \left (3 x
               \left (\sqrt {x} + \sqrt {x + 1} \right) - 2
               \sqrt {x + 1} \right) - 1} {2 \sqrt {(-x + 1) x}
    \sqrt {x + 1} \sqrt {2 (x - 1) \sqrt {x + 1} x^{3/2} + 2
        x^3 - x^2 - x + 1}}$$
</s>