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