# Sin X − X Cos X X Sin X + Cos X - Mathematics

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

#### Solution

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

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 13 | Page 44