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प्रश्न
Differentiate each of the following from first principle:
\[e^\sqrt{2x}\]
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उत्तर
\[\frac{d}{dx}\left( e^\sqrt{2x} \right) = \lim_{h \to 0} \frac{e^\sqrt{2(x + h)} - e^\sqrt{2x}}{h}\]
\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - e^\sqrt{2x}}{2x + 2h - 2x}\]
\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x} \left( e^\sqrt{2x + 2h} - \sqrt{2x} - 1 \right)}{\left( \sqrt{2x + 2h} \right)^2 - \left( \sqrt{2x} \right)^2}\]
\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]
\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)} \lim_{h \to 0} \frac{1}{\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]
\[ = 2 e^\sqrt{2x} \left( 1 \right)\frac{1}{2\sqrt{2x}}\]
\[ = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]
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