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

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Differentiate each of the following from first principle:

$\frac{\cos x}{x}$

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#### Solution

$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\frac{\cos \left( x + h \right)}{x + h} - \frac{\cos x}{x}}{h}$
$= \lim_{h \to 0} \frac{x \cos \left( x + h \right) - \left( x + h \right) \cos x}{h x \left( x + h \right)}$
$= \lim_{h \to 0} \frac{x \left( \cos x \cos h - \sin x \sin h \right) - x \cos x - h \cos x}{h x \left( x + h \right)}$
$= \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h - x \cos x - h \cos x}{h x \left( x + h \right)}$
$= \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h - h \cos x}{h x \left( x + h \right)}$
$= x\cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h}$
$= x \cos x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h} \left[ \because \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}} = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} = 1 \times 1, i . e . 1 \right]$
$= - x \cos x \lim_{h \to 0} \frac{h}{2} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h}$
$= - x \cos x \times 0 - \sin x \left( 1 \right)\frac{1}{x} - \frac{\cos x}{x}\frac{1}{x}$
$= 0 - \frac{\sin x}{x} - \frac{\cos x}{x^2}$
$= - \frac{\sin x}{x} - \frac{\cos x}{x^2}$
$= \frac{- x \sin x - \cos x}{x^2}$

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 3.03 | Page 26

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