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