if the


$(\log x)^{\log x}, x>1$


Let $y=(\log x)^{\log x}$

Taking logarithm on both the sides, we obtain

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

Differentiating both sides with respect to x, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \cdot \log (\log x)]$

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

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

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

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


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