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प्रश्न
Differentiate the following function from first principles \[e^\sqrt{\cot x}\] .
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उत्तर
\[\text{Let} f\left( x \right) = e^\sqrt{\cot x} \]
\[ \Rightarrow f\left( x + h \right) = e^\sqrt{\cot\left( x + h \right)} \]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^\sqrt{\cot\left( x + h \right)} - e^\sqrt{\cot x}}{h}\]
\[ = \lim_{h \to 0} \frac{e^\sqrt{\cot x} \left( e^{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} - 1 \right)}{h}\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \left(\frac{e^{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} - 1}{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} \right) \times \left( \frac{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}}{h} \right)\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\left( \sqrt{\cot\left( x + h \right)} - \sqrt{\cot x} \right)}{h} \times \frac{\sqrt{\cot\left( x + h \right)} + \sqrt{\cot x}}{\sqrt{\cot\left( x + h \right)} + \sqrt{\cot x}} \left[ \because \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \text{ and rationalizing the numerator } \right]\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right) - \cot x}{h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\frac{\cot\left( x + h \right)\cot x + 1}{\cot\left( x - x - h \right)}}{h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)} \left[ \because \cot\left( A - B \right) = \frac{\cot A\cot B + 1}{\cot B - \cot A} \right]\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right)\cot x + 1}{\cot\left( - h \right) \times h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = - e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right)\cot x + 1}{\left( \frac{h}{\tanh} \right)\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = \frac{e^\sqrt{\cot x} \times \left( \cot^2 x + 1 \right)}{2\sqrt{\cot x}} \left[ \because \lim_{x \to 0} \frac{\tan x}{x} = 1 \right]\]
\[ = - \frac{e^\sqrt{\cot x} \times {cosec}^2 x}{2\sqrt{\cot x}} \left[ \because \left( 1 + \cot^2 x \right) = {cosec}^2 x \right]\]
\[ \therefore \frac{d}{dx}\left( e^\sqrt{cot x} \right) = - \frac{e^\sqrt{\cot x} \times {cosec}^2 x}{2\sqrt{\cot x}}\]
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