f(x)=ln(ln(1-x))
Find f '(x)
The derivative of $\displaystyle \ln(u(x))$ is $\displaystyle \frac{u'(x)}{u(x)}$
Here $\displaystyle u(x) = \ln(1-x) = \ln(v(x))$ where $\displaystyle v(x) = 1-x$
$\displaystyle u'(x) = \frac{v'(x)}{v(x)} = \frac{-1}{1-x}$
Therefore $\displaystyle f'(x) = \frac{\frac{-1}{1-x}}{\ln(1-x)} = -\frac{1}{(1-x)\ln(1-x)}$