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))))#

:)