I'm bad at math, but this is because d/dx ln(x) = x^-1, while d/dx x^k = kx^(-1+k), so as long as k > 0 the derivative of the latter is larger, right?
Well sure but you can also just graph them or compare Taylor series (in radius of convergence, appropriate cut, etc)
Well, the question here was just "is ln(x) sublinear or not?". You can answer that question with much less work: the second derivative is always negative, so the function must be sublinear. Any function that grows linearly must have a second derivative of zero.