Question
Differentiate the following functions with respect to x:
$3\text{e}^{-3\text{x}}\log(1+\text{x})$

Answer

Consider $\text{y}=3\text{e}^{-3\text{x}}\log(1+\text{x})$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\big[3\text{e}^{-3\text{x}}\log(1+\text{x})\big]$
$\frac{\text{dy}}{\text{dt}}=3\Big(\text{e}^{-3\text{x}}\frac{1}{1+\text{x}}+\log(1+\text{x})\big(-3\text{e}^{-3\text{x}}\big)\Big)$
$=3\Big(\frac{\text{e}^{-3\text{x}}}{1+\text{x}}-3\log(1+\text{x})\Big)$
The solution is,
$=3\text{e}^{-3\text{e}}\Big(\frac{1}{1+\text{x}}-3\log(1-\text{x})\Big)$

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