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