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

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

By taking log on both sides , we get

log y $=$ log x log (log x) ...(i)
Differentiating (i) on both sides w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \log x \cdot \cfrac{1}{{\log x}} \cdot \cfrac{1}{x} + \log (\log x)\cfrac{1}{x} = \cfrac{1}{x} \cdot [1 + \log (\log x)]$

therefore, $\cfrac{{dy}}{{dx}} = {(\log x)^{\log x}} \cdot \cfrac{1}{x} \cdot [1 + \log (\log x)].$