Question

Differentiate log_x 3 ?

Answer 1

\[\text{ Let y } = \log_x 3\]

\[ \Rightarrow y = \frac{\log3}{\log x} \left[ \because \log_a b = \frac{\log b}{\log a} \right]\]

\[\text{Differentiate it with respect to x we get}, \]

\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\log3}{\log x} \right)\]

\[ = \log3\frac{d}{dx} \left( \log x \right)^{- 1} \]

\[ = \log3 \times \left[ - 1 \left( \log x \right)^{- 2} \right]\frac{d}{dx}\left( \log x \right) \left[\text{ using chain rule} \right]\]

\[ = - \frac{\log3}{\left( \log x \right)^2} \times \frac{1}{x}\]

\[ = - \left( \frac{\log3}{\log x} \right)^2 \times \frac{1}{x} \times \frac{1}{\log3}\]

\[ = - \frac{1}{x\log3 \left( \log_3 x \right)^2} \left[ \because \frac{\log b}{\log a} = \log_a b \right]\]

\[So, \frac{d}{dx}\left( \log_x 3 \right) = - \frac{1}{x\log3 \left( \log_3 x \right)^2}\]