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Question
Differentiate \[3 e^{- 3x} \log \left( 1 + x \right)\] ?
Solution
\[\text{Let } y = 3 e^{- 3x} \log\left( 1 + x \right)\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = 3\frac{d}{dx}\left[ e^{- 3x} \log\left( 1 + x \right) \right] \]
\[ = 3\left\{ e^{- 3x} \frac{1}{1 + x} + \log\left( 1 + x \right)\left( - 3 e^{- 3x} \right) \right\} \left[ \text{Using product rule and chain rule} \right]\]
\[ = 3\left\{ \frac{e^{- 3x}}{1 + x} - 3 e^{- 3x} \log\left( 1 + x \right) \right\}\]
\[ = 3 e^{- 3x} \left\{ \frac{1}{1 + x} - 3 \log\left( 1 + x \right) \right\}\]
\[So, \frac{d}{dx}\left[ 3 e^{- 3x} \log\left( 1 + x \right) \right] = 3 e^{- 3x} \left\{ \frac{1}{1 + x} - 3 \log\left( 1 + x \right) \right\}\]
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