Question

If $y={\cos }^{-1}\left(2x\sqrt{1-x^2}\right)$, then find $\frac{dy}{dx}$

Answer 1

Let $y={\cos }^{-1}\left(2x\sqrt{1-x^2}\right)$
Put $x=\sin \theta$
$\therefore \theta ={\sin }^{-1}x$
$\Rightarrow y={\cos }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2}\theta \right)$
$\Rightarrow y={\cos }^{-1}\left(2\sin \theta \cos \theta \right)$
$\Rightarrow y={\cos }^{-1}\left(\sin 2\theta \right)$
$\Rightarrow y={\cos }^{-1}\left[\cos (\frac{\pi }{2}-2\theta )\right]$
$\Rightarrow y=\frac{\pi }{2}-2\theta$
$\Rightarrow y=\frac{\pi }{2}-2{\sin }^{-1}x$
$\Rightarrow \frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$