Differentiate each of the following from first principle:

\[e^\sqrt{2x}\]

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#### Solution

\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]

\[\frac{d}{dx}\left( e^\sqrt{2x} \right) = \lim_{h \to 0} \frac{e^\sqrt{2(x + h)} - e^\sqrt{2x}}{h}\]

\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - e^\sqrt{2x}}{2x + 2h - 2x}\]

\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x} \left( e^\sqrt{2x + 2h} - \sqrt{2x} - 1 \right)}{\left( \sqrt{2x + 2h} \right)^2 - \left( \sqrt{2x} \right)^2}\]

\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]

\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)} \lim_{h \to 0} \frac{1}{\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]

\[ = 2 e^\sqrt{2x} \left( 1 \right)\frac{1}{2\sqrt{2x}}\]

\[ = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]

\[\frac{d}{dx}\left( e^\sqrt{2x} \right) = \lim_{h \to 0} \frac{e^\sqrt{2(x + h)} - e^\sqrt{2x}}{h}\]

\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - e^\sqrt{2x}}{2x + 2h - 2x}\]

\[ = 2 \lim_{h \to 0} \frac{e^\sqrt{2x} \left( e^\sqrt{2x + 2h} - \sqrt{2x} - 1 \right)}{\left( \sqrt{2x + 2h} \right)^2 - \left( \sqrt{2x} \right)^2}\]

\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]

\[ = 2 e^\sqrt{2x} \lim_{h \to 0} \frac{e^\sqrt{2x + 2h} - \sqrt{2x} - 1}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)} \lim_{h \to 0} \frac{1}{\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}\]

\[ = 2 e^\sqrt{2x} \left( 1 \right)\frac{1}{2\sqrt{2x}}\]

\[ = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]

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