Question

If $y=x{\sin }^{-1}x+\sqrt{1-x^2}$, prove that $\frac{dy}{dx}={\sin }^{-1}x$

Answer 1

$\mathrm{We}\mathrm{have},y=x{\sin }^{-1}x+\sqrt{1-x^2}$
Differentiate it with respect to x,
$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left[x{\sin }^{-1}x+\sqrt{1-x^2}\right] \\ =\frac{d}{dx}\left(x{\sin }^{-1}x\right)+\frac{d}{dx}\left(\sqrt{1-x^2}\right) \\ =\left[x\frac{d}{dx}{\sin }^{-1}x+{\sin }^{-1}x\frac{d}{dx}\left(x\right)\right]+\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}\left(1-x^2\right) \\ =\left[\frac{x}{\sqrt{1-x^2}}+{\sin }^{-1}x\right]-\frac{2x}{2\sqrt{1-x^2}} \\ =\frac{x}{\sqrt{1-x^2}}+{\sin }^{-1}x-\frac{x}{\sqrt{1-x^2}} \\ ={\sin }^{-1}x \end{gathered}$$