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Question
Differentiate (log x)x with respect to log x ?
Solution
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = x\left( \frac{1}{\log x} \right)\frac{d}{dx}\left( \log x \right) + \log\log x\left( 1 \right)\]
\[ \Rightarrow \frac{du}{dx} = u\left[ \frac{x}{\log x}\left( \frac{1}{x} \right) + \log \log x \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( \log x \right)^x \left[ \frac{1}{\log x} + \log \log x \right] . . \left( i \right)\]
\[\text { Again, let v } = \log x\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{x} . . . \left( ii \right)\]
\[\text { Dividing equation } \left( i \right) \text { by } \left( ii \right), \text { we get }\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\left( \log x \right)^x \left[ \frac{1}{\log x} + \log \log x \right]}{\frac{1}{x}}\]
\[ \Rightarrow \frac{du}{dv} = \frac{\left( \log x \right)^x \left[ \frac{1 + \log x\left( \log \log x \right)}{\log x} \right]}{\frac{1}{x}}\]
\[ \Rightarrow \frac{du}{dv} = x \left( \log x \right)^{x^{- 1}} \left( 1 + \log x \times \log \log x \right)\]
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