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