find the dervative of .

Possible derivation:
d/dx(log^x(x))
Express log^x(x) as a power of e: log^x(x) = e^(log(log^x(x))) = e^(x log(log(x))):
  =  d/dx(e^(x log(log(x))))
Using the chain rule, d/dx(e^(x log(log(x)))) = ( de^u)/( du) ( du)/( dx), where u = x log(log(x)) and ( d)/( du)(e^u) = e^u:
  =  (d/dx(x log(log(x)))) e^(x log(log(x)))
Express e^(x log(log(x))) as a power of log(x): e^(x log(log(x))) = e^(log(log^x(x))) = log^x(x):
  =  log(x)^x d/dx(x log(log(x)))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = log(log(x)):
  =  log(log(x)) d/dx(x)+x d/dx(log(log(x))) log^x(x)
The derivative of x is 1:
  =  log^x(x) (x (d/dx(log(log(x))))+1 log(log(x)))
Using the chain rule, d/dx(log(log(x))) = ( dlog(u))/( du) ( du)/( dx), where u = log(x) and ( d)/( du)(log(u)) = 1/u:
  =  log^x(x) (log(log(x))+(d/dx(log(x)))/(log(x)) x)
The derivative of log(x) is 1/x:
  =  log^x(x) (log(log(x))+(1/x x)/(log(x)))
Simplify the expression:
Answer: |   =  log^x(x) (1/(log(x))+log(log(x)))

y = (ln(x) )^x      take the ln of both sides

ln(y)  =  ln [ln(x)]^x

ln (y)  = x ln [ln (x) ]     now.....differentiate both sides

y' / y   = ln [ln (x) ]   +  x [  (1/x) / ln(x) ]    simplify

y' / y = ln [ ln (x) ] + 1/ [ln (x) ]    multiply both sides by y

y' = y [ ln [ln (x)  + 1/ [ln (x) ]      and substitute  [  ln(x) ]^x   for y

y'  = [ln (x)]^x [ ln [ln(x)] +  1 / ln (x) ]

Ordering of functions  from  slowest growing to fastest growing  as x  →  infinity

 e^(x/2)        e^x       [ln (x)]^x         x^x