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