Posted by daria on Sunday, March 30, 2008 at 1:40am.
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
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