Question
Differentiate the following function with respect to x:

$\text{x}^3\text{e}^\text{x}\cos\text{x}$

Answer

Let $\text{u}=\text{x}^3;\text{v}=\text{e}^\text{x};\text{w}=\cos\text{x}$

Then, $\text{u}'=\text{3x}^2;\text{v}'=\text{e}^\text{x};\text{w}'=-\sin\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uv}'\text{w}+\text{uv}\text{w}'$

$\frac{\text{d}}{\text{dx}}(\text{x}^3\text{e}^\text{x}\cos\text{x})=\text{3x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^3\text{e}^\text{x}\cos\text{x}+\text{x}^3\text{e}^\text{x}(-\sin\text{x})$

$=\text{x}^2\text{e}^\text{x}(\text{3}\cos\text{x}+\text{x}\cos\text{x}-\text{x}\sin\text{x})$

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