Question

Differentiate (log x)^x with respect to log x

Answer 1

$\mathrm{Let}u={\left(\log x\right)}^x$
Taking log on both sides,
$$\begin{gathered} \log u=\log {\left(\log x\right)}^x \\ \Rightarrow \log u=x\log \left(\log x\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}\left\{\log \left(\log x\right)\right\}+\log \left(\log x\right)\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=x\left(\frac{1}{\log x}\right)\frac{d}{dx}\left(\log x\right)+\log \log x\left(1\right) \\ \Rightarrow \frac{du}{dx}=u\left[\frac{x}{\log x}\left(\frac{1}{x}\right)+\log \log x\right] \\ \Rightarrow \frac{du}{dx}={\left(\log x\right)}^x\left[\frac{1}{\log x}+\log \log x\right]..\left(\mathrm{i}\right) \\ \mathrm{Again},\mathrm{let}v=\log x \\ \Rightarrow \frac{dv}{dx}=\frac{1}{x}...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{by}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{{\left(\log x\right)}^x\left[{\displaystyle \frac{1}{\log x}}+\log \log x\right]}{{\displaystyle \frac{1}{x}}} \\ \Rightarrow \frac{du}{dv}=\frac{{\left(\log x\right)}^x\left[{\displaystyle \frac{1+\log x\left(\log \log x\right)}{\log x}}\right]}{{\displaystyle \frac{1}{x}}} \\ \Rightarrow \frac{du}{dv}=x{\left(\log x\right)}^{x^{-1}}\left(1+\log x\times \log \log x\right) \end{gathered}$$