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Question
Differentiate logx 3 ?
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Solution
\[\text{ Let y } = \log_x 3\]
\[ \Rightarrow y = \frac{\log3}{\log x} \left[ \because \log_a b = \frac{\log b}{\log a} \right]\]
\[\text{Differentiate it with respect to x we get}, \]
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\log3}{\log x} \right)\]
\[ = \log3\frac{d}{dx} \left( \log x \right)^{- 1} \]
\[ = \log3 \times \left[ - 1 \left( \log x \right)^{- 2} \right]\frac{d}{dx}\left( \log x \right) \left[\text{ using chain rule} \right]\]
\[ = - \frac{\log3}{\left( \log x \right)^2} \times \frac{1}{x}\]
\[ = - \left( \frac{\log3}{\log x} \right)^2 \times \frac{1}{x} \times \frac{1}{\log3}\]
\[ = - \frac{1}{x\log3 \left( \log_3 x \right)^2} \left[ \because \frac{\log b}{\log a} = \log_a b \right]\]
\[So, \frac{d}{dx}\left( \log_x 3 \right) = - \frac{1}{x\log3 \left( \log_3 x \right)^2}\]
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