Question

Find $\frac{dy}{dx}$

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

Answer 1

$y=si{n}^{-1}\left(2x\sqrt{1-x^2}\right)$

Let $x=sin\theta$

Therefore,

$y=si{n}^{-1}(2sin\theta \sqrt{1-si{n}^2\theta )}$

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

$y=si{n}^{-1}sin2\theta$

$y=2\theta$

$y=2si{n}^{-1}x$

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