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