Solution
= Solution
For a real-valued expression, $\ln x>0$ requires $x>1$. Applying the <chain rule> twice gives
$$
\boxed{\frac d{dx}\ln(\ln x)=\frac1{\ln x}\frac1x=\frac1{x\ln x},\qquad x>1.}
$$
= Solution
For a real-valued expression, $\ln x>0$ requires $x>1$. Applying the <chain rule> twice gives
$$
\boxed{\frac d{dx}\ln(\ln x)=\frac1{\ln x}\frac1x=\frac1{x\ln x},\qquad x>1.}
$$