Differentiate the following functions with respect to x :

Let $y=3 e^{-3 x} \log (1+x)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[3 e^{-3 x} \log (1+x)\right]$

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

We have (uv)' $=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{d y}{d x}=3\left[\log (1+x) \frac{d}{d x}\left(e^{-3 x}\right)+e^{-3 x} \frac{d}{d x}[\log (1+x)]\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}}$ and $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

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

$\Rightarrow \frac{d y}{d x}=3\left[-3 e^{-3 x} \log (1+x) \frac{d}{d x}(x)+\frac{e^{-3 x}}{1+x}\left(\frac{d}{d x}(1)+\frac{d}{d x}(x)\right)\right]$

However, $\frac{d}{d x}(x)=1$ and derivative of a constant is 0 .

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3\left[-3 \mathrm{e}^{-3 \mathrm{x}} \log (1+\mathrm{x}) \times 1+\frac{\mathrm{e}^{-3 \mathrm{x}}}{1+\mathrm{x}}(0+1)\right]$

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{e}^{-3 \mathrm{x}}\left[\frac{1}{1+\mathrm{x}}-3 \log (1+\mathrm{x})\right]$

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