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Question:

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

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}=0$

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

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