The derivative of $(\log \mathrm{x})^{\mathrm{x}}$ with respect to $\log \mathrm{x}$ is
The derivative of $(\log \mathrm{x})^{\mathrm{x}}$ with respect to $\log \mathrm{x}$ is
- $(\log x)^x\left[\frac{1}{\log x} \log (\log x)\right]$
- $(\log )^x\left[\log x+\frac{1}{\log (\log x)}\right]$
- $x(\log )^x\left[\frac{1}{\log x}+\log (\log x)\right]$
- $x(\log )^x\left[\log x+\frac{1}{\log (\log x)}\right]$
Solution
Let $\mathrm{u}=(\log \mathrm{x})^{\mathrm{x}}$
$\begin{aligned}
& \therefore \log \mathrm{u}=\mathrm{x} \log \left[\log (\mathrm{x})^{\mathrm{x}}\right] \\
& \therefore \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{x}}{\log (\mathrm{x})} \times \frac{1}{\mathrm{x}}+\log (\log \mathrm{x})=\log (\log \mathrm{x})+\frac{1}{\log \mathrm{x}} \\
& \therefore \frac{\mathrm{du}}{\mathrm{dx}}=(\log \mathrm{x})^{\mathrm{x}}\left[\frac{1}{\log \mathrm{x}}+\log (\log \mathrm{x})\right]
\end{aligned}$
Let $v=\log x \Rightarrow \frac{d v}{d x}=\frac{1}{x}$
$\therefore \frac{d u}{d v}=(\log x)^x(x)\left[\frac{1}{\log x}+\log (\log x)\right]$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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