derivatives-x^(x^x)

f(x)=x^(x^x)=y
In(y)=x^x.In(x)
In(In(y))=In(x^x.In(x))
In(In(y))=x.In(x)+In (In(x))......(1)
Method 1
Using chain rules:
dy/dx.1/In(y).1/y=(x.1/x+1.In(x))+In(x)/x.1/x .....(2)
dy/dx=In(y).y(1+In(x)+In(x)/(x^2))
dy/dx=x^x.In(x).x^(x^x)(1+In(x)+In(x)/(x^2))
Method 2
In(In(y))=x.In(x)+In (In(x))
y=f(x)=e^{e^[x.In(x)+In (In(x))]} ->f(g(u(x)))
dy/dx= f'(g(u(x)))+g'(u(x))+u'(x)=
e^{e^[x.In(x)+In (In(x))]}+e^[x.In(x)+In (In(x))]+d/dx x.In(x)+In(In(x))
using(2):
e^{e^[x.In(x)+In (In(x))]}+e^[x.In(x)+In (In(x))]+(x.1/x+1.In(x))+In(x)/x.1/x
e^{e^[x.In(x)+In (In(x))]}+e^[x.In(x)+In (In(x))]+(1+In(x)+In(x)/(x^2))
Method 3
A bit similar to method 1
See:http://www.youtube.com/watch?v=N5kkwVoAtkc&feature=player_embedded#!
dy/dx=x^(x^x)[x^x(In(x)+1)In(x)+x(x-1)]

Damn, they drive me nuts, please help!!!