Solve this

Given $y=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}}}{\mathrm{x}+1}\left[\frac{1}{2}(\mathrm{x})^{-\frac{1}{2}}-\frac{1}{2}(\mathrm{x})^{-\frac{3}{2}}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}}}{2(\mathrm{x}+1)}\left[\frac{1}{\mathrm{x}^{\frac{1}{2}}}-\frac{1}{\mathrm{x}^{\frac{3}{2}}}\right]$

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

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

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

Thus, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-1}{2 \mathrm{x}(\mathrm{x}+1)}$