What is derivative of x^[ln(x)]

I have a test tomorrow and I don't understand this, please help!

Recall ,if f(x)=ln[g(x)], then f'(x)=g'(x)/g(x).

The clearest way to me to do this is write this as
ln[f(x)]=ln(x)*ln(x)=[ln(x)]^2.(that is begin by taking
log of both sides). Then, by the above comment,
f'(x)/f(x)=2ln(x)/x (using the chain rule).
Hence
f'(x)=f(x)*2*ln(x)/x=2*ln(x)*x^[ln(x)]…

Hope that helps.

x^[ln(x)]=e^(lnx)^2

u=lnx

du=1/x


e^u^2=2ueu^2

2lnxe^(lnx)^2/x