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# X Sin X 1 + Cos X - Mathematics

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$\frac{x \sin x}{1 + \cos x}$

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#### Solution

$\text{ Let } u = x \sin x; v = 1 + \cos x$
$\text{ Then }, u' = x \cos x + \sin x; v' = - \sin x$
$\text{ Using the quotient rule }:$
$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}$
$\frac{d}{dx}\left( \frac{x \sin x}{1 + \cos x} \right) = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) - x \sin x\left( - \sin x \right)}{\left( 1 + \cos x \right)^2}$
$= \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x \sin^2 x}{\left( 1 + \cos x \right)^2}$
$= \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x \left( 1 - \cos^2 x \right)}{\left( 1 + \cos x \right)^2}$
$= \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x\left( 1 + \cos x \right)\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)^2}$
$= \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x + x - x\cos x \right)}{\left( 1 + \cos x \right)^2}$
$= \frac{\left( 1 + \cos x \right)\left( x + \sin x \right)}{\left( 1 + \cos x \right)^2}$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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#### APPEARS IN

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

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