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

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    $\begingroup$ 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. $\endgroup$ – A-Level Student Oct 27 '20 at 19:09
  • $\begingroup$ @A-levelStudent, I tried this approach but it was involving too much calculation, so I thought some substitution may simplify it. $\endgroup$ – 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:

$$\begin{align} \cos\alpha=x\implies-\sin\alpha\frac{\mathrm d\alpha}{\mathrm dx}&=1\\[1ex]\frac{\mathrm d\alpha}{\mathrm dx}&=-\csc\alpha\\[1ex] \frac{\mathrm d\alpha}{\mathrm dx}&=-\frac1{\sqrt{1-x^2}}\end{align}$$

$$\begin{align}\cos\beta=\sqrt x\implies-\sin\beta\frac{\mathrm d\beta}{\mathrm dx}&=\frac1{2\sqrt x}\\[1ex]\frac{\mathrm d\beta}{\mathrm dx}&=-\frac{\csc\beta}{2\sqrt x}\\[1ex]\frac{\mathrm d\beta}{\mathrm dx}&=-\frac1{2\sqrt x\sqrt{1-x^2}}\end{align}$$


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