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