(2) If $x>1$, then the value of $x^{\ln \ln x}-(\ln x)^{\ln x}$ is $(\quad)$.
(A) Positive
(B) Zero
(C) Negative
(D) Any of the above is possible
(2) B Hint: Let $y=\ln x$, then $x=\mathrm{e}^{y}$, the original expression $=\left(\mathrm{e}^{y}\right)^{\ln y}-y^{y}=0$.