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Question
Differentiate \[\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)\] ?
Solution
\[\text{ Let } y = \log\left( 3x + 2 \right) - x^2 \log\left( 2x - 1 \right)\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ \log\left( 3x + 2 \right) - x^2 \log\left( 2x - 1 \right) \right]\]
\[ = \frac{d}{dx}\log\left( 3x + 2 \right) - \frac{d}{dx}\left\{ x^2 \log\left( 2x - 1 \right) \right\}\]
\[ = \frac{1}{\left( 3x + 2 \right)}\frac{d}{dx}\left( 3x + 2 \right) - \left[ x^2 \frac{d}{dx}\log\left( 2x - 1 \right) + \log\left( 2x - 1 \right)\frac{d}{dx}\left( x^2 \right) \right] \left[ \text{Using product rule and chain rule} \right]\]
\[ = \frac{3}{3x + 2} - \frac{2 x^2}{\left( 2x - 1 \right)} - 2x \log\left( 2x - 1 \right)\]
\[So, \frac{d}{dx}\left[ \log\left( 3x + 2 \right) - x^2 \log\left( 2x - 1 \right) \right] = \frac{3}{3x + 2} - \frac{2 x^2}{\left( 2x - 1 \right)} - 2x \log\left( 2x - 1 \right)\]
\[\]
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