# Differentiate of the Following from First Principle: X Cos X - Mathematics

Differentiate of the following from first principle:

x cos x

#### Solution

$\left( x \right) \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\left( x + h \right) \cos \left( x + h \right) - x \cos x}{h}$
$= \lim_{h \to 0} \frac{\left( x + h \right)\left( \cos x \cos h - \sin x \sin h \right) - x \cos x}{h}$
$= \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h - x \cos x}{h}$
$= \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h}{h}$
$= x \cos x \lim_{h \to 0} \frac{\left( \cos h - 1 \right)}{h} - x \sin x \lim_{h \to 0} \frac{\sin h}{h} + \cos x \lim_{h \to 0} \cos h + \sin x \lim_{h \to 0} \sin h$
$= x \cos x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - x \sin x \left( 1 \right) + \cos x \left( 1 \right) + \sin x \left( 0 \right)$
$= x\cos x \lim_{h \to 0} \frac{- h}{2} - x \sin x \left( 1 \right) + \cos x \left( 1 \right) + \sin x \left( 0 \right)$
$= - x \cos x \left( 0 \right) - x \sin x + \cos x$
$= - x \sin x + \cos x$



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 30.2 | Q 2.1 | Page 25