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Question
\[\frac{x + \cos x}{\tan x}\]
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Solution
\[\text{ Let } u = x + \cos x; v = \tan x\]
\[\text{ Then }, u' = 1 - \sin x; v' = \sec^2 x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{x + \cos x}{\tan x} \right) = \frac{\tan x\left( 1 - \sin x \right) - \left( x + \cos x \right) \sec^2 x}{\tan^2 x}\]
\[\]
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