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