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Question
\[\frac{4x + 5 \sin x}{3x + 7 \cos x}\]
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Solution
\[\text{ Let } u = 4x + 5 \sin x; v = 3x + 7 \cos x\]
\[\text{ Then }, u' = 4 + 5 \cos x; v' = 3 - 7 \sin x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{4x + 5 \sin x}{3x + 7 \cos x} \right) = \frac{\left( 3x + 7 \cos x \right)\left( 4 + 5 \cos x \right) - \left( 4x + 5 \sin x \right)\left( 3 - 7 \sin x \right)}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{12x + 15 x \cos x + 28 \cos x + 35 \cos^2 x - 12x + 28 x \sin x - 15 \sin x + 35 \sin^2 x}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{15 x \cos x + 28 x \sin x + 28 \cos x15 \sin x + 35\left( \sin^2 x + \cos^2 x \right)}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{15 x \cos x + 28 x \sin x + 28 \cos x15 \sin x + 35}{\left( 3x + 7 \cos x \right)^2}\]
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