I want to find the derivative of $y = \log_b(\log_b(x))$
I am going to let $u = \log_b(x)$ so that $y = \log_b(u)$. BY Chain Rule, I get
$\frac{dy}{dx} =\frac{dy}{du} \frac{du}{dx}$
$\frac{dy}{du} = \frac{1}{u\ln(b)} = \frac{1}{\log_b(x)\ln(b)}$
$\frac{du}{dx} = \frac{1}{x\ln(b)}$
$\frac{dy}{dx} =\frac{dy}{du} \frac{du}{dx} =\frac{1}{\log_b(x)\ln(b)} \frac{1}{x\ln(b)} = \frac{1}{\ln^2(b)x\log_b(x)}$
The answer I got from Wolframalpha is http://www.wolframalpha.com/input/?i=D[log[log%28x%29]%2Cx]&a=*C.D-_*Function.dflt-&a=*FunClash.log-_*Log10.Log-
Which they only have one $\ln(b)$.
Note that they have $b = 10$ in this case.