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प्रश्न
Differentiate the following functions from first principles \[e^\sqrt{2x}\].
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उत्तर
\[\text{ Let } f\left( x \right) = e^\sqrt{2x} \]
\[ \Rightarrow f\left( x + h \right) = e^\sqrt{2\left( x + h \right)} \]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^{{}^\sqrt{2\left( x + h \right)}} - e^{{}^\sqrt{2x}}}{h}\]
\[ = \lim_{h \to 0} e^\sqrt{2x} \left[ \frac{e^\sqrt{2\left( x + h \right)} - \sqrt{2x} - 1}{h} \right]\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \left[ \frac{e^\sqrt{2\left( x + h \right)} - \sqrt{2x} - 1}{\sqrt{2\left( x + h \right)} - \sqrt{2x}} \right] \times \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h} \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h} \times \frac{\sqrt{2\left( x + h \right)} + \sqrt{2x}}{\sqrt{2\left( x + h \right)} + \sqrt{2x}} \] [Rationalising the numerator]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2\left( x + h \right) - 2x}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2x + 2h - 2x}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)} \]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2h}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2}{\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]
\[\text{ Hence }, \frac{d}{dx}\left( e^\sqrt{2x} \right) = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]
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