Question

$\frac{d}{dx}{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=.......|x|>\frac{1}{\sqrt{2}}$

Answer 1

Let $y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)|x|>\frac{1}{\sqrt{2}}$

Let $x=\sin \theta$

$\Rightarrow 1-x^2=1-{\sin }^{\theta }={\cos }^2\theta$

$\Rightarrow y={\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)$

$={\sin }^{-1}\left(2\sin \theta |\cos \theta |\right)$

$={\sin }^{-1}\left(\sin 2\theta \right)$

Now $|x|>\frac{1}{\sqrt{2}}$

$\Rightarrow \theta \in [\frac{-\pi }{2},\frac{-\pi }{4})\cup (\frac{\pi }{4},\pi /2]$

$\Rightarrow 2\theta \in [-\pi ,-\frac{\pi }{2})\cup (\pi /2,\pi ]$

$\Rightarrow y={\sin }^{-1}\left(\sin 2\theta \right)=\{\begin{array}{c}-\pi -2\theta 2\theta \in [-\pi ,\frac{-\pi }{2})\\ +\pi -2\theta 2\theta \in (\pi /2,\pi ]\end{array}$

$\Rightarrow y=\{\begin{array}{c}-\pi -2{\sin }^{-1}xx\in [-1,-\frac{1}{\sqrt{2}})\\ \pi -2{\sin }^{-1}xx\in (\frac{1}{\sqrt{2}},1]\end{array}$

$\Rightarrow \frac{dy}{dx}=-2\frac{1}{\sqrt{1-x^2}}$

$\Rightarrow \frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$