Two derivatives of $y=\arcsin 2x\sqrt {1-x^2}$? $y=\arcsin 2x\sqrt {1-x^2} ,\frac{-1}{\sqrt2}\lt x \lt \frac{1}{\sqrt2}$
I. Substituting $x=\sin m$
$y=\arcsin (2\sin m \cos m)=\arcsin (\sin2m)=2m=2\arcsin x$
$\Rightarrow y'=\frac{2}{\sqrt{1-x^2}}$
II. If you substitute $x=\cos m$, you get $y=2m=2 \arccos x$
$\Rightarrow$ $y'=\frac{-2}{\sqrt {1-x^2}}$
So, my question is, how can there be two different derivatives of the same function?
Also, another question I have is, in every question the domain of the function is given, is there any use of that in the process of finding the derivative?
 A: If $\arccos x=y,$  
using the definition of Principal values, $0\le y\le\pi$
$$\arcsin2x\sqrt{1-x^2}=\arcsin(\sin2y)=\begin{cases}2y=2\arccos x&\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff1\ge x\ge\dfrac1{\sqrt2} \\ 
\pi-2y=2\arcsin x & \mbox{if }-\dfrac\pi2\le\pi-2y\le\dfrac\pi2\iff-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}\\-2\pi+2y& \mbox{if }-1\le x\le-\dfrac1{\sqrt2}\end{cases}$$
A: Hmmm, this is a tricky problem. But let's re-simplify the function again:  
Substituting $x=\sin m$
$y=\arcsin (2\sin m \cos m)=\arcsin (\sin2m)$ all good!  
However, $\arcsin (\sin2m)=2m $ not $\forall m \in \mathbb R$, but $\forall m \in [-\frac{\pi}{4},\frac{\pi}{4}]$. Here's where the domain of $y$ comes into play. The function is defined on $[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}]$, so you can see for yourself that the smallest interval on which $m$ satisfies $x=\sin m$ should be $[-\frac{\pi}{2},\frac{\pi}{2}]\supset[-\frac{\pi}{4},\frac{\pi}{4}]$.
Thus, there's a mistake in your simplification process leading to two conflicting derivatives. You can't simply cancel out arcsin and sin on that interval. But, nevertheless, you can define a new function  $f(x)=\arcsin(\sin(2x))$ and look for its explicit formula (it's rather easy, a piecewise one) and then have  
$y=f(2m)=f(2\arcsin x)$   
Then differentiate with chain rule on each piece. No problem should occur then!  
$y'=f'(2\arcsin x)\cdot2(\arcsin x)'$ (And you will see f'(x) is a step function)
Same logic with $x=\cos m$ part.   
Wish you understand the problem now :D
