Is it possible for the derivative of a function to grow arbitrarily faster than the function itself?

Question

We know that there exist some functions $f(x)$ such that their derivative $f'(x)$ is strictly greater than the function itself. for example the function $5^x$ has a derivative $5^x\ln(5)$ which is greater than $5^x$. Exponential functions in general are known to be proportional to their derivatives. The question I have is whether it is possible for a function to grow "even faster" than this. To be more precise let's take the ratio $f'(x)/f(x)$ for exponential functions this ratio is a constant. For most elementary functions we care about, this ratio usually tends to 0. But are there functions for which this ratio grows arbitrarily large? If so, is there an upper limit for how large the ratio $f'(x)/f(x)$ can grow? I also ask a similar question for integrals.

Answer 1

Consider the differential equation $$ \frac{f'}{f} = g $$ where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get $$f(x) = e^{e^x-1} $$ The ratio $f'/f$ grows arbitrarily large.

Answer 2

You can always try functions of the form $f(x) = e^{g(x)}$, where $g(x)$ is an antiderivative of a function with large growth. For example, if $f(x) = e^{e^x}$, then $$\frac{f'(x)}{f(x)} = e^x,$$ which is, of course, exponential.

Answer 3

Even $f(x)=\frac 1x$ has derivative $\frac {-1}{x^2}$ and second derivative $\frac 2{x^3}$ which grow arbitrarily faster than $f(x)$ as $x \to 0$. The ratio $\frac {f'(x)}{f(x)}=-\frac 1x$ which is not bounded as $x \to 0$. The important message is that derivatives accentuate short range changes, so if you have a function that changes quickly in a short distance the derivative is large.

I don't understand what "a similar question for integrals" means. Integrals are smoothing functions, so the integral of a function can't grow faster than the function times the length of the integral.