Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$ 
Derivative of $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$

Put $\sqrt{x}=\sin\alpha,\;\sqrt{a}=\sin\beta$
$$
y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]=\sin^{-1}\bigg[\sqrt{x(1-a)}-\sqrt{a(1-x)}\bigg]\\
=\sin^{-1}\bigg[\sin\alpha|\cos\beta|-|\cos\alpha|\sin\beta\bigg]
$$
My reference gives the solution $\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$, but is it a complete solution ?
How do I proceed further with my attempt ?

Using
$$ (\arcsin u)'=\frac1{\sqrt{1-u^2}}u' $$
one has
\begin{eqnarray}
\frac{dy}{dx}&=&\frac{1}{\sqrt{1-(\sqrt{x-ax}-\sqrt{a-ax})^2}}(\sqrt{x-ax}-\sqrt{a-ax})'\\
&=&\frac1{\sqrt{1-(x-ax)-(a-ax)+2\sqrt{a(1-a)x(1-x)}}}\bigg[\frac{\sqrt{1-a}}{2\sqrt x}+\frac{\sqrt a}{2\sqrt{1-x}}\bigg]\\
&=&\frac{1}{\sqrt{(1-a)(1-x)+ax+2\sqrt{a(1-a)x(1-x)}}}\cdot\frac{\sqrt{(1-a)(1-x)}+\sqrt{ax}}{2\sqrt{x(1-x)}}\\
&=&\frac{1}{2\sqrt{x(1-x)}},
\end{eqnarray}
where
$$ 1-(x-ax)-(a-ax)=(1-a)(1-x)+ax $$
is used.
