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))))#
:)
