1. The problem is to simplify the inequality $\frac{d}{dx} \ln(\ln x) > 0$. 2. Recall the derivative rule for a composite function: if $y = \ln(u)$, then $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$. 3. Here, $u = \ln x$, so $\frac{du}{dx} = \frac{1}{x}$. 4. Therefore, the derivative is: $$\frac{d}{dx} \ln(\ln x) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}.$$ 5. The inequality becomes: $$\frac{1}{x \ln x} > 0.$$ 6. For a fraction to be positive, its numerator and denominator must have the same sign. 7. The numerator is 1, which is always positive. 8. Thus, the denominator $x \ln x$ must be positive: $$x \ln x > 0.$$ 9. Since $x > 0$ (domain of $\ln x$), the sign depends on $\ln x$. 10. $\ln x > 0$ when $x > 1$. 11. Therefore, the solution to the inequality is: $$x > 1.$$