<s>[INST] Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$  ? 
EDIT: So many people have answered that the domain would be $[-1,1]$ but my book says that its $[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}].$ 
Can anyone explain how are those restrictions made in the given formulas? 
 [/INST] You should see the domain is at first restricted by the square root $\sqrt{1-x^2}$ so that we must have $|x|\le 1$. Also note that $g(x) = 2x\sqrt{1-x^2}$ is an odd function, so we can just analyse for $x> 0$.
$$g'(x) = 2 \sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}} = \frac{2-4x^2}{\sqrt{1-x^2}}$$
Giving maxima at $x = \tfrac{1}{\sqrt{2}}$. So $g(\tfrac{1}{\sqrt{2}}) = 1$ and $g(0) = 0$. So we have shown that $0 \le g(x) \le 1$ for $x \in [0,1]$. Similarly we show $0 \ge g(x) \ge -1$ for $x \in [-1,0]$
So required domain is $[-1,1]$
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