In a problem, we have to differentiate : $$ y = \arcsin (2 x \sqrt{1 - x^2}),$$ given: $$-{1/(\sqrt2)} < x <{1/(\sqrt2)}$$.
My approach :
Put $x = \cos (p)$, so y reduces to:- $y = 2p = 2 \arccos (x) \implies$ ${dy}/{dx}$ $ = -{2}/{\sqrt{1-x^2}}$
But the solution given is without the minus sign, so I am doing something wrong here but don't know what.
So please tell me what I am doing wrong and how to correct it.
$$-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$$
If $x=\cos P, \dfrac\pi4\le P\le\dfrac{3\pi}4$ as $0\le\arccos x\le\pi$
$\implies\dfrac\pi2\le2P\le\dfrac{3\pi}2$
But as $-\dfrac\pi2\le\arcsin(\sin2P)\le\dfrac\pi2$
$\arcsin(\sin2P)=\pi-2P=\pi-2\arccos x$