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Question
\[\frac{x \tan x}{\sec x + \tan x}\]
Solution
\[\text{ Let } u = x \tan x; v = \sec x + \tan x\]
\[\text{ Then }, u' = x \sec^2 x + \tan x; v' = \sec x \tan x + \sec^2 x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{x\tan x}{\sec x + \tan x} \right) = \frac{\left( \sec x + \tan x \right)\left( x \sec^2 x + \tan x \right) - x \tan x\left( \sec x \tan x + \sec^2 x \right)}{\left( \sec x + \tan x \right)^2}\]
\[ = \frac{x \sec^3 x + x \sec^2 x\tan x + \sec x \tan x + \tan^2 x - x \sec x \tan^2 x - x \tan x \sec^2 x}{\left( \sec x + \tan x \right)^2}\]
\[ = \frac{\left( \sec x + \tan x \right)\left( x \sec^2 x + \tan x \right) - x \tan x \sec x\left( \sec x + \tan x \right)}{\left( \sec x + \tan x \right)^2}\]
\[ = \frac{x \sec^2 x + \tan x - x \tan x \sec x}{\sec x + \tan x}\]
\[ = \frac{x \sec x\left( \sec x - \tan x \right) + \tan x}{\sec x + \tan x}\]
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