Question

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

Answer 1

$y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2}$
$\therefore \frac{dy}{dx}=\frac{d}{dx}[{\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2}]$
$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left({\sin }^{-1}x\right)+\frac{d}{dx}\left({\sin }^{-1}\sqrt{1-x^2}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2}{)}^2}}.\frac{d}{dx}\left(\sqrt{1-x^2}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}.\frac{1}{2\sqrt{1-x^2}}.\frac{d}{dx}(1-x^2)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{2x\sqrt{1-x^2}}(-2x)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0$