Question
Differentiate the following function with respect to $\text{x}:$
$(\text{x}^3+\text{x}^2+1)\sin\text{x}$
$(\text{x}^3+\text{x}^2+1)\sin\text{x}$
Then,
$\text{u}'=\text{3x}^2+\text{2x};\text{v}'=\cos\text{x}$Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(\text{x}^3+\text{x}^2+1)\sin\text{x}]$
$=(\text{x}^3+\text{x}^2+1)\cos\text{x}+(\text{3x}^2+\text{2x})\sin\text{x}$
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