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

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.

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.

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.