The first derivative of ln (ln x) is dy/dx = 1/(x*ln x)
The second derivative is:
d^2y/dx^2 = [-ln x/x^2 - (1/x^2)]/(ln x)^2
= -1/(x^2 ln x) -1/(x ln x)^2
x d^2y/dx^2 = -1/(x ln x) - (1/x)/lnx^2
It seems to fit the differential equation
Find the second derivative of y=loge(logex)
Show that y=loge(logex) is a solution of the equation
xd^2y/dx^2 + x(dy/dx)^2 + dy/dx=o
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