Find the $\frac{\mathrm{d} y}{\mathrm{d} x}$ of $ \arcsin (x \sqrt{1-x} - \sqrt{x}\sqrt{1-x^2})$ I tried to solve this question first by trigonometric substitution like putting $x =\sin^2{\theta}$ but this complicated the calculation. Is there any simple technique to solve this question. The answer to this problem is $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x-x^2}}$
 A: I wouldn't call this much of a shortcut, but it does make the computation a bit cleaner.
Let $\cos\alpha=x$ and $\sin\alpha=\sqrt{1-x^2}$. So
$$x\sqrt{1-x}-\sqrt x\sqrt{1-x^2}=\cos\alpha\sqrt{1-\cos\alpha}-\sqrt{\cos\alpha}\sin\alpha$$
Now, let $\sin\beta=\sqrt{1-\cos\alpha}$ and $\cos\beta=\sqrt{\cos\alpha}$. Notice that $\sin^2\beta+\cos^2\beta=1$ holds. So really, we have
$$\begin{align}
x\sqrt{1-x}-\sqrt x\sqrt{1-x^2}&=\cos\alpha\sin\beta-\cos\beta\sin\alpha\\[1ex]
&=\sin(\beta-\alpha)
\end{align}$$
which reduces the original function to $y=\beta-\alpha$ (on an appropriate domain). Thus
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm d\beta}{\mathrm dx}-\frac{\mathrm d\alpha}{\mathrm dx}$$
Compute the remaining derivatives using $\cos\beta=\sqrt{\cos\alpha}=\sqrt x$ and $\cos\alpha=x$ along with the chain rule:
$$\begin{align}
\cos\alpha=x\implies-\sin\alpha\frac{\mathrm d\alpha}{\mathrm dx}&=1\\[1ex]\frac{\mathrm d\alpha}{\mathrm dx}&=-\csc\alpha\\[1ex]
\frac{\mathrm d\alpha}{\mathrm dx}&=-\frac1{\sqrt{1-x^2}}\end{align}$$
$$\begin{align}\cos\beta=\sqrt x\implies-\sin\beta\frac{\mathrm d\beta}{\mathrm dx}&=\frac1{2\sqrt x}\\[1ex]\frac{\mathrm d\beta}{\mathrm dx}&=-\frac{\csc\beta}{2\sqrt x}\\[1ex]\frac{\mathrm d\beta}{\mathrm dx}&=-\frac1{2\sqrt x\sqrt{1-x^2}}\end{align}$$
