Question

If $y=\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2}$ then prove that

$\frac{dy}{dx}=\frac{x\sqrt{1-x^2}+{\sin }^{-1}x(1+x^2)-2x(1-x^2)}{\sqrt{1-x^2}(1-x^2)}$

Answer 1

Take $u=\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}$

$\frac{du}{dx}=\frac{\sqrt{1-x^2}\frac{d}{dx}\left(x{\sin }^{-1}x\right)-x{\sin }^{-1}x\frac{d}{dx}\left(\sqrt{1-x^2}\right)}{(\sqrt{1-x^2}{)}^2}$

$=\frac{\sqrt{1-x^2}[\frac{x}{\sqrt{1-x^2}}+{\sin }^{-1}x]-x{\sin }^{-1}x\left(\frac{-2x}{\sqrt{1-x^2}}\right)}{(1-x^2)}$

$=\frac{x+\sqrt{1-x^2}{\sin }^{-1}x+\frac{2x^2{\sin }^{-1}x}{\sqrt{1-x^2}}}{(1-x^2)}$

$=\frac{x\sqrt{1-x^2}+(1-x^2){\sin }^{-1}x+2x^2{\sin }^{-1}x}{(1-x^2)\sqrt{1-x^2}}$

$=\frac{x\sqrt{1-x^2}+{\sin }^{-1}x+x^2{\sin }^{-1}x}{(1-x^2)\sqrt{1-x^2}}$

$=\frac{x\sqrt{1-x^2}+{\sin }^{-1}x+x^2{\sin }^{-1}x}{(1-x^2)\sqrt{1-x^2}}$

Let $v=log\sqrt{1-x^2}\Rightarrow \frac{dv}{dx}=\frac{(-2x)}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$\frac{dy}{dx}=\frac{x\sqrt{1-x^2}+{\sin }^{-1}x(1+x^2)-2x(1-x^2)}{\sqrt{1-x^2}(1-x^2)}$.