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### Answers (1)

Answer: $\log x^{x-1}(1+\log x \log \log x)$

Hint: Differentiate by  $\log ^{x}\left ( x \right )$

Given: $(\log x)^{x}$

Solution:  Let $y=(\log x)^{x}$

Taking log both sides

$\log y=x \log (\log x)$

Differentiate w.r.t $x$,

$\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{d}{d x}\{\log (\log x)\}+\log (\log x)\left(\frac{d}{d x}\right)(x)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=x \frac{1}{\log x} \frac{d}{d x}(\log x)+\log (\log x)(1)$

\begin{aligned} &\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{x}{\log x}\left(\frac{1}{x}\right)+\log (\log x) \\\\ &\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{\log x}+\log (\log x) \\\\ &\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{\log x}+\log (\log x)\right] \end{aligned}

Now, put value of   $y=(\log x)^{x}$

$\Rightarrow \frac{d y}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]$

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