Question

Find $\frac{dy}{dx}$in the following questions:

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

Answer 1

Let $si{n}^{-1}\left(2x\sqrt{1-x^2}\right)\Rightarrow y=si{n}^{-1}\left(2sin\theta \sqrt{1-si{n}^2\theta }\right)(\because 1-si{n}^2\theta =co{s}^2\theta )$

$\Rightarrow y=si{n}^{-1}\left(2sin\theta cos\theta \right)=si{n}^{-1}\left(sin2\theta \right)$

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

Differentiating both sides w.r.t. x, we get

$\Rightarrow \frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}(\because si{n}^{-1}x=\frac{1}{\sqrt{1-x^2}})$