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X 1 + Tan X - Mathematics

\[\frac{x}{1 + \tan x}\] 

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Solution

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

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 6 | Page 44
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