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Question
\[\frac{\sin x - x \cos x}{x \sin x + \cos x}\]
Solution
\[\text{ Let } u = \sin x - x \cos x; v = x \sin x + \cos x\]
\[\text{ Then }, u' = \cos x + x \sin x - \cos x; v' = x \cos x + \sin x - \sin x\]
\[ = x \sin x = x \cos x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{\sin x - x \cos x}{x \sin x + \cos x} \right) = \frac{\left( x \sin x + \cos x \right)x \sin x - \left( \sin x - x \cos x \right)x \cos x}{\left( x \sin x + \cos x \right)^2}\]
\[ = \frac{x^2 \sin^2 x + x \cos x \sin x - x \cos x \sin x + x^2 \cos^2 x}{\left( x \sin x + \cos x \right)^2}\]
\[ = \frac{x^2 \left( \sin^2 x + \cos^2 x \right)}{\left( x \sin x + \cos x \right)^2}\]
\[ = \frac{x^2}{\left( x \sin x + \cos x \right)^2}\]
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