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प्रश्न
Differentiate of the following from first principle:
e3x
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उत्तर
\[\frac{d}{dx}\left( e^{3x} \right) = \lim_{h \to 0} \frac{e^{3(x + h)} - e^{3x}}{h}\]
\[ = \lim_{h \to 0} \frac{e^{3x} e^{3h} - e^{3x}}{h}\]
\[ = \lim_{h \to 0} \frac{e^{3x} \left( e^{3h} - 1 \right)}{3h}\]
\[ = 3 e^{3x} \lim_{h \to 0} \frac{e^{3h} - 1}{3h}\]
\[ = 3 e^{3x} \left( 1 \right)\]
\[ = 3 e^{3x}\]
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