# 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 $$\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x-x^2}}$$

• We know that the derivative of $f(g(x))$ is $f'(g(x))g'(x)$. Try applying that a few times to your expression. If that doesn't help tell me and I'll try to gve an answer. – A-Level Student Oct 27 '20 at 19:09
• @A-levelStudent, I tried this approach but it was involving too much calculation, so I thought some substitution may simplify it. – Ranjeet Bahadur Oct 27 '20 at 19:16

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}