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Sec X − 1 Sec X + 1 - Mathematics

\[\frac{\sec x - 1}{\sec x + 1}\] 

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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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 25 | Page 44
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