# Solve this problem

Question:

If $y=\sqrt{a^{2}-x^{2}}$, prove that $y \frac{d y}{d x}+x= 0$

Solution:

Given $y=\sqrt{a^{2}-x^{2}}$

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}}\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ [using chain rule]

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

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}}[0-2 x]$

$\Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 \sqrt{a^{2}-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a^{2}-x^{2}}}$

But, $y=\sqrt{a^{2}-x^{2}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\mathrm{y}}$

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

$\therefore y \frac{d y}{d x}+x=0$

Thus, $y \frac{d y}{d x}+x=0$