Question

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

Answer 1

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