<s>[INST] Why does simplifying $\arcsin(2x\sqrt{1-x^2})$ two ways give different results? 
Find derivative of
  $y=\arcsin(2x \sqrt{1-x^2}) $ in domain $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

If you put $x=\sin\theta$ then
$$ y= \arcsin(2sin(\theta) \sqrt{1- sin^2 ( \theta)})$$
$$y= \arcsin( sin2\theta)$$
$$y= 2\theta$$
$$y = 2 \arcsin(x)$$
But, if you put $x=\cos\theta$ then , again,
$$y=2\theta$$
But,  resubstituting
$$ y= 2 \arccos(x)$$
But derivatives of both are different.

Now where's the mistake? Is it something related to the original domain I took?

 [/INST] There is nothing wrong with your algebra. It’s just that you need to take extra care when dealing with the domain. Note that $\arcsin(\sin \theta=\theta$ only when $-\frac{\pi}{2} \le \theta\le \frac{\pi}{2}$. 
Substituting $x=\sin\theta$ gives $$y= \arcsin(2\sin\theta|\cos\theta|)=\arcsin|\sin 2\theta| $$ Now, $$-\frac{1}{\sqrt 2} \lt \sin\theta \lt \frac{1}{\sqrt 2} \\ \implies -\frac{\pi}{4} \le \theta\le \frac{\pi}{4} \\\implies -\frac{\pi}{2} \lt2\theta \lt \frac{\pi}{2}$$ and we can safely say that $$y=2\theta=2\arcsin x$$
But when $x=\cos \theta$, we have $$ -\frac{1}{\sqrt 2} \lt \cos\theta \lt \frac{1}{\sqrt 2} \\ \implies \frac{\pi}{4} \lt \theta \lt \frac{3\pi}{4} \\ \implies \frac{\pi}{2} \lt 2\theta \lt \frac{3\pi}{2}$$ This time, we need to shift by $\pi$ so that $$-\frac{\pi}{2}\lt 2\theta-\pi\lt\frac{\pi}{2}$$So, $$ y =\arcsin(\sin 2\theta) =\arcsin\left(-\sin(2\theta-\pi)\right) = \pi-2\arccos x \\ = 2\left(\frac{\pi}{2}-\arccos x\right) \\ = 2\arcsin x$$
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