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