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

Differentiate each of the following from first principle:

sin x + cos x

#### Solution

$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\sin \left( x + h \right) + cos \left( x + h \right) - \sin x - \cos x}{h}$
$= \lim_{h \to 0} \frac{\sin \left( x + h \right) - \sin x}{h} + \lim_{h \to 0} \frac{\cos \left( x + h \right) - \cos x}{h}$
$\text{ We have }:$
$sin C-sin D= 2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)$
$And, cos C- \cos D = - 2 \sin\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)$
$= \lim_{h \to 0} \frac{2 \cos \left( \frac{2x + h}{2} \right) \sin \frac{h}{2}}{h} + \lim_{h \to 0} \frac{- 2 \sin \left( \frac{2x + h}{2} \right) \sin \frac{h}{2}}{h}$
$= 2 \lim_{h \to 0} \cos \left( \frac{2x + h}{2} \right) \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2} - 2 \lim_{h \to 0} \sin \left( \frac{2x + h}{2} \right) \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2}$
$= 2 \cos x \times \frac{1}{2} - 2 \sin x \times \frac{1}{2}$
$= \cos x - \sin x$

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 3.06 | Page 26