Find the points of local maximum and minimun of the function: $$f(x)=\sin^{-1}(2x\sqrt{1-x^2})~~~~;~~x\in (-1,1)$$ I know $$f'(x)=-\frac{2}{\sqrt{1-x^2}}$$
How to find the local maximum and minimum? I have drawn the fig and seen the points of local maximum and minimum. But how to find then analytically?
As suggested in the comment using my answer here in Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$,
$$ \arcsin(2x\sqrt{1-x^2}) =\begin{cases}2\arcsin x \;\;;-\dfrac1{\sqrt2}\le x\le \dfrac1{\sqrt2}\iff-\dfrac\pi4\le\arcsin x\le\dfrac\pi4\\ \pi - 2\arcsin x\;\;; \dfrac1{\sqrt2}< x\le 1\iff\dfrac\pi4<\arcsin x\le\dfrac\pi2\\ -\pi -2\arcsin x\;\;;-1< x \le-\dfrac1{\sqrt2}\iff-\dfrac\pi2\le\arcsin x<-\dfrac\pi4\end{cases} $$
Can you take it from here?
First of all, a bit of theory.
Suppose you have a function $F$ defined over the interval $(a,b)$ and increasing thereon. Suppose also that $G$ takes on values in $(a,b)$. Then, in order to find the points where $F(G(x))$ has a local maximum or minimum, it's sufficient to find the points where $G$ has a local maximum or minimum. This is because $F(G(x_1))<F(G(x_2))$ if and only if $G(x_1)<G(x_2)$, as $F$ is increasing.
The arcsine is an increasing function, so you just need to find the local maxima and minima of $g(x)=x\sqrt{1-x^2}$. You should already have worked out that $-1\le 2x\sqrt{1-x^2}\le 1$ for $x\in[-1,1]$ when you have discussed the domain of $f$. With notation as before, $G(x)=2x\sqrt{1-x^2}$ and $F(t)=\arcsin t$.
Since $$ G'(x)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=\frac{1-2x^2}{\sqrt{1-x^2}} $$ you find that:
$\lim_{x\to-1^+}G'(x)=-\infty=\lim_{x\to1^-}G'(x)$
$G'(x)$ vanishes at $-1/\sqrt{2}$ and at $1/\sqrt{2}$, being positive in the interval $(-1/\sqrt{2},1/\sqrt{2})$.
Now you should be able to end: $G$ has a local minimum at $-1/\sqrt{2}$ and a local maximum at $1/\sqrt{2}$.
If you include $-1$ and $1$ in the domain of $f$, then $-1$ is a local maximum point and $1$ a local minimum point.
As a side note, $f$ is not differentiable at $\pm1/\sqrt{2}$, but this is irrelevant.
Your computation of the derivative is wrong: it should be $$ f'(x)=\frac{1}{\sqrt{1-(2x\sqrt{1-x^2})^2}}\frac{1-2x^2}{\sqrt{1-x^2}}= \frac{1-2x^2}{|1-2x^2|}\frac{1}{\sqrt{1-x^2}} $$ Note that $\sqrt{(1-2x^2)^2}=|1-2x^2|$.
From this you could derive the same conclusions as before. The function is increasing where $1-2x^2>0$ and decreasing where $1-2x^2<0$.
Hint: We know that $$\arcsin x + \arcsin y = \arcsin (x\sqrt{1-y^2} + y\sqrt {1-x^2}) $$ If $x=y $, then $$2\arcsin x = \arcsin (2x\sqrt {1-x^2}) $$
Hope you can take it from here.
