Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

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

Differentiate each of the following from first principle:

$\frac{\sin x}{x}$

#### Solution

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

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 30.2 | Q 3.02 | Page 26