# Solve this

Question:

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

Solution:

Given $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}$

On differentiating $y$ with respect to $x$, we get

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

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

We have $(u v)^{\prime}=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{1-x^{2}}\left(\sin ^{-1} x \frac{d}{d x}(x)+x \frac{d}{d x}\left(\sin ^{-1} x\right)\right)-\left(x \sin ^{-1} x\right) \frac{d}{d x}\left(\left(1-x^{2}\right)^{\frac{1}{2}}\right)}{1-x^{2}}$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$ and $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}$

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

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

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

However, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)=2 \mathrm{x}$ and derivative of a constant is 0 .

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{1-\mathrm{x}^{2}} \sin ^{-1} \mathrm{x}+\mathrm{x}+\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \times \mathrm{x}}{1-\mathrm{x}^{2}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=\sqrt{1-x^{2}} \sin ^{-1} x+x+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=x+\sqrt{1-x^{2}} \sin ^{-1} x+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\left(\sqrt{1-x^{2}}\right)^{2} \sin ^{-1} x+x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\left(1-x^{2}\right) \sin ^{-1} x+x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\left(1-x^{2}+x^{2}\right) \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\Rightarrow\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$

But, $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} \Rightarrow \frac{y}{x}=\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\therefore\left(1-\mathrm{x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}}$

Thus, $\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}$