Question

Find $\frac{dy}{dx}$, if $y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2},0<x<1$

Answer 1

Let $y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2}$

Substituting $x=\sin \theta$

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

$y=\theta +{\sin }^{-1}\sqrt{{\cos }^2\theta }$

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

$\because \cos \theta =\sin (\frac{\pi }{2}-\theta )$

$y=\theta +{\sin }^{-1}\left(\sin (\frac{\pi }{2}-\theta )\right)$

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

$y=\theta +(\frac{\pi }{2}-\theta )$

$y=\frac{\pi }{2}$

Differentiating 𝑦 w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{d\left(\frac{\pi }{2}\right)}{dx}$

$\frac{dy}{dx}=0$