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Question
\[\frac{1 + 3^x}{1 - 3^x}\]
Solution
\[\text{ Let } u = 1 + 3^x ; v = 1 - 3^x \]
\[\text{ Then }, u' = 3^x \log 3; v' = - 3^x \log 3\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{1 + 3^x}{1 - 3^x} \right) = \frac{\left( 1 - 3^x \right) 3^x \log 3 - \left( 1 + 3^x \right)\left( - 3^x \log 3 \right)}{\left( 1 - 3^x \right)^2}\]
\[ = \frac{3^x \log 3 - 3^{2x} \log 3 + 3^x \log 3 + 3^{2x} \log 3}{\left( 1 - 3^x \right)^2}\]
\[ = \frac{2 . 3^x \log 3}{\left( 1 - 3^x \right)^2}\]
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