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