If $\mathrm{y}=\sqrt{\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}}}$, prove that $2 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}$.
Given $y=\sqrt{x}+\frac{1}{\sqrt{x}}$
On differentiating $y$ with respect to $x$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}}}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})+\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})^{\frac{1}{2}}+\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})^{-\frac{1}{2}}$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2}(x)^{\frac{1}{2}-1}+\left(-\frac{1}{2}\right)(x)^{-\frac{1}{2}-1}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}(\mathrm{x})^{-\frac{1}{2}}-\frac{1}{2}(\mathrm{x})^{-\frac{3}{2}}$c
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left[\frac{1}{\mathrm{x}^{\frac{1}{2}}}-\frac{1}{\mathrm{x}^{\frac{3}{2}}}\right]$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\frac{1}{\sqrt{x}}-\frac{1}{x \sqrt{x}}\right]$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\frac{x-1}{x \sqrt{x}}\right]$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-1}{2 \mathrm{x} \sqrt{\mathrm{x}}}$
$\Rightarrow 2 x \frac{d y}{d x}=\frac{x-1}{\sqrt{x}}$
$\Rightarrow 2 x \frac{d y}{d x}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}$
$\therefore 2 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}$
Thus, $2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$