Differentiate the function with respect to $x$:
$(\log x)^{\log x}, x > 1$

Let $y = (\log x)^{\log x}$.
Taking the logarithm on both sides,we get:
$\log y = \log x \cdot \log(\log x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[\log x] \cdot \log(\log x) + \log x \cdot \frac{d}{dx}[\log(\log x)]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \cdot \log(\log x) + \log x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = \frac{\log(\log x)}{x} + \frac{1}{x}$.
Therefore,the derivative is:
$\frac{dy}{dx} = y \left[ \frac{\log(\log x) + 1}{x} \right] = (\log x)^{\log x} \left[ \frac{1 + \log(\log x)}{x} \right]$.