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