We need to rewrite the expression via chain rule.
First: ln(x) = u.
Now, we have ln(ln(u)) as our original function.
Second: ln(u) = z.
Now, we have ln(z) as our original function.
We now have to derive ln(z).
(dln(z))/dx = (z')/z, where z' is the derivative of z.
However, we know z: it's ln(u).
So, (dln(z))/dx = ([ln(u)]')/(ln(u)), where [ln(u)]' stands for the derivative of ln(u).
But we know that the derivative of a lnf(x) is (f'(x))/f(x), so we can rewrite:
(dln(z))/dx = ((u')/u)/ln(u).
Again, we know u. It's ln(x), isn't it?
Substituting...
(dln(z))/dx = ([ln(x)]')/(lnx)/(ln(ln(x)))
Going part by part, now.
[ln(x)]' = 1/x, thus ([ln(x)]')/(lnx) = 1/(x.ln(x)).
Going back to our original derivation again,
(dln(ln(ln(x))))/dx = 1/(x.ln(x))/(ln(ln(x))) = 1/(x.ln(x).(ln(ln(x))))
:)
