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प्रश्न
\[\frac{e^x - \tan x}{\cot x - x^n}\]
उत्तर
\[\text{ Let } u = e^x - \tan x; v = \cot x - x^n \]
\[\text{ Then }, u' = e^x - \sec^2 x; v' = - \cos e c^2 x - n x^{n - 1} \]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{e^x - \tan x}{cot x - x^n} \right) = \frac{\left( \cot x - x^n \right)\left( e^x - \sec^2 x \right) - \left( e^x - \tan x \right)\left( - \cos e c^2 x - n x^{n - 1} \right)}{\left( \cot x - x^n \right)^2}\]
\[ = \frac{\left( \cot x - x^n \right)\left( e^x - \sec^2 x \right) + \left( e^x - \tan x \right)\left( \cos e c^2 x + n x^{n - 1} \right)}{\left( \cot x - x^n \right)^2}\]
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