Derivative of x^x^x

Y=x^(x^x)
logy=x^xlogx
log(logy)=logx^x + logx
log(logy)=xlogx +logx
1/ylogy * dy/dx= 1+logx + 1/x
dy/dx= x^(x^x) * x^xlogx[1+logx+1/x]
Where right answer is – dy/dx= x^x^x * x^x[(1+logx)logx +1/x]

wher am i wrong.. ?

When I work it out, I get

y = x^(x^x)
log y = log x^(x^x)
log y = x^x * log x
log (log y) = log(x^x) + log (log x) <-------------Here's the problem
log (log y) = x log x + log (log x)

d/dx (log (log y)) = 1/log y * d/dx (log y)
d/dx (log (log y)) = 1/log y * 1/y * dy/dx
d/dx (log (log y)) = 1/(y log y) * dy/dx

d/dx (x log x + log (log x)) = d/dx (x log x) + d/dx (log (log x))
d/dx (x log x + log (log x)) = 1 log x + x * 1/x + 1/log x * d/dx (log x)
d/dx (x log x + log (log x)) = log x + 1 + 1/log x * 1/x
d/dx (x log x + log (log x)) = log x + 1 + 1/(x log x)

1/(y log y) * dy/dx = log x + 1 + 1/(x log x)
dy/dx = y log y * (log x + 1 + 1/(x log x))
dy/dx = x^(x^x) log x^(x^x) * (log x + 1 + 1/(x log x))
dy/dx = x^(x^x) x^x log x * (log x + 1 + 1/(x log x))
dy/dx = x^(x^x) x^x [(log x + 1) log x + 1/x]

When you took the 2nd log of both sides, you should have wound up with log(log x) instead of just log x. I hope that helps!