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

# Differentiate of the Following from First Principle: Cos ( X − π 8 ) - Mathematics

Differentiate  of the following from first principle:

$\cos\left( x - \frac{\pi}{8} \right)$

#### Solution

$\frac{d}{dx}\left( f\left( x \right) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$\frac{d}{dx}\left( \cos \left( x - \frac{\pi}{8} \right) \right) = \lim_{h \to 0} \frac{\cos \left( x + h - \frac{\pi}{8} \right) - \cos \left( x - \frac{\pi}{8} \right)}{h}$
$We know:$
$\cos C - \cos D = - 2 \sin \left( \frac{C + D}{2} \right) \sin \left( \frac{C - D}{2} \right)$
$= \lim_{h \to 0} \frac{- 2 \sin \left( \frac{x + h - \frac{\pi}{8} + x - \frac{\pi}{8}}{2} \right) \sin \left( \frac{x + h - \frac{\pi}{8} - x + \frac{\pi}{8}}{2} \right)}{h}$
$= \lim_{h \to 0} \frac{- 2 \sin \left( \frac{2x + h - \frac{\pi}{4}}{2} \right) \sin \left( \frac{h}{2} \right)}{h}$
$= - 2 \lim_{h \to 0} \sin \left( \frac{2x + h - \frac{\pi}{4}}{2} \right) \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \times \frac{1}{2}$
$= - 2 \sin \left( x - \frac{\pi}{8} \right) \times \frac{1}{2}$
$= - \sin \left( x - \frac{\pi}{8} \right)$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 2.08 | Page 25