Find the derivative of (log x)^log x,x>1

Let $y=(\log x)^{\log x}$. Taking log in both sides we get $\log y=\log x\log(\log x)$. Differentiating both sides w.r.to x we get $1/y\frac{dy}{dx}=\frac{1+\log(\log x)}{x}$. Hence $\frac{dy}{dx}=y(\frac{1+\log(\log x)}{x})=(\log x)^{\log x}(\frac{1+\log(\log x)}{x})$