Question
Differentiate the following function with respect to $\text{x}$:
$\text{x}^3\text{e}^\text{x}$
$\text{x}^3\text{e}^\text{x}$
Then,
$\text{u}'=3\text{x}^2;\text{v}'=\text{e}^\text{x}$Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^3\text{e}^\text{x})=\text{x}^3\text{e}^\text{x}+\text{e}^\text{x}(3\text{x}^2)$
$=\text{x}^2\text{e}^\text{x}(\text{x}+3)$
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$\frac{\text{ax}^2+\text{bx}+\text{c}}{\text{px}^2+\text{qx}+\text{r}}$