Differentiate the following functions with respect to x :

Let $y=3^{x \log x}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(3^{x \log \mathrm{x}}\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}^{\mathrm{x}}\right)=\mathrm{a}^{\mathrm{x}} \log \mathrm{a}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3^{\mathrm{x} \log \mathrm{x}} \log 3 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \log \mathrm{x})$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3^{\mathrm{x} \log \mathrm{x}} \log 3 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \times \log \mathrm{x})$

Recall that $(u v)^{\prime}=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{d y}{d x}=3^{x \log x} \log 3\left[\log x \frac{d}{d x}(x)+x \frac{d}{d x}(\log x)\right]$

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

$\Rightarrow \frac{d y}{d x}=3^{x \log x} \log 3\left[\log x \times 1+x \times \frac{1}{x}\right]$

$\Rightarrow \frac{d y}{d x}=3^{x \log x} \log 3[\log x+1]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=(1+\log \mathrm{x}) 3^{\mathrm{xlog} \mathrm{x}} \log 3$

Thus, $\frac{d}{d x}\left(3^{x \log x}\right)=(1+\log x) 3^{x \log x} \log 3$