4 ms·
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?
by sterlind 4y ago
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?
- selimthegrim 4y agoWell sure but you can also just graph them or compare Taylor series (in radius of convergence, appropriate cut, etc)
- thaumasiotes 4y agoWell, 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.