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प्रश्न
Differentiate of the following from first principle:
\[\cos\left( x - \frac{\pi}{8} \right)\]
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उत्तर
\[\frac{d}{dx}\left( f\left( x \right) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[\frac{d}{dx}\left( \cos \left( x - \frac{\pi}{8} \right) \right) = \lim_{h \to 0} \frac{\cos \left( x + h - \frac{\pi}{8} \right) - \cos \left( x - \frac{\pi}{8} \right)}{h}\]
\[We know:\]
\[\cos C - \cos D = - 2 \sin \left( \frac{C + D}{2} \right) \sin \left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{- 2 \sin \left( \frac{x + h - \frac{\pi}{8} + x - \frac{\pi}{8}}{2} \right) \sin \left( \frac{x + h - \frac{\pi}{8} - x + \frac{\pi}{8}}{2} \right)}{h}\]
\[ = \lim_{h \to 0} \frac{- 2 \sin \left( \frac{2x + h - \frac{\pi}{4}}{2} \right) \sin \left( \frac{h}{2} \right)}{h}\]
\[ = - 2 \lim_{h \to 0} \sin \left( \frac{2x + h - \frac{\pi}{4}}{2} \right) \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \times \frac{1}{2}\]
\[ = - 2 \sin \left( x - \frac{\pi}{8} \right) \times \frac{1}{2}\]
\[ = - \sin \left( x - \frac{\pi}{8} \right)\]
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