Asymptotic behavior and derivatives

[maze]

If you take the derivative of, say, x^k, you get kx^k-1, a function that grows more slowly as x approaches infinity. On the other hand, if you take the derivative of x^x, you get x^x( ln(x) + 1), a function that grows faster than x^x. In fact, if you do this experiment with most standard "nice" functions, you will find that that the derivative makes fastly growing things grow even faster, and slowly growing things grow even slower, with e^x as a happy medium whose derivative grows at the same rate as itself.

On the other hand, it is not hard to construct "bad" functions that simultaneously grow and wiggle, for which this pattern does not hold.

Has anyone looked into this before? I would be interested to know what class of functions behave like this (ie: the derivative is asymptotically smaller than the original function if the original function is asymptotically smaller than e^x and the derivative is asymptotically larger if the function is asymptotically larger than e^x).

 

[olliemath]

Interesting.. it looks like the sort of question that might've been studied by victorian mathematicians.. like you say it's quite easy to construct even monotonically increasing functions for which the patturn is broken, so I don't think there'll be a general result.

If we call f(x) superexponential if e^{ax}\in o(f(x)) for all a>0 then it is reasonably easy to show that derivatives of superexp. functions are superexp. and those of subexp. are subexp. etc.

The question of the derivatives having higher order seems much more subtle..
Let us suppose that f(x) is positive on some interval [\alpha,\infty) and for each a>0, e^{ax}/f(x)\to 0 and the limit is eventually monotonic. So by e.g. the MVT the derivative is negative for all x>N_a say.
That is
\frac{ae^{ax}f(x)-e^{ax}f'(x)}{f(x)^2}<0 ~\Rightarrow~\frac{f(x)}{f'(x)}<a^{-1}
for all x>N_a. For large a, a^{-1} is small, hence f(x)/f'(x)\to 0.

One could replace this monotonicity condition with e.g. the condition that the derivative of q_a(x)=e^{ax}/f(x) is bounded by a suitable function p_a(x), for each a which tends to 0 fast enough at infinity.

These two conditions seem a little intractible. I would think that the class of convex functions would be one for which this all holds true, and there is a wealth of literature on them.. still, it's an intriguing question.. might have more of a play with it at some point.