If I have a differentiable $L^2$ function $\psi:\mathbb R\rightarrow \mathbb C$ which is normalised $$ \int |\psi(x)|^2\;\text d x = 1 $$ can I say anything about the order of $$ \int |\psi'(x)|^2\;\text d x\quad \text ? $$
I don't want to keep it as general as possible. More the other way round: What shall I assume to be able to say something? For example we can assume that $|\psi|$ has its single maximum at $x=0$ and then decays rapidly. Furthermore we can say that $\psi$ is an hermitian function.
I made an example with a gaussian with a simple phase $$ \psi(x)=\frac{1}{(2\pi\sigma^2)^{1/4}}\text e^{-\frac{x^2}{4\sigma^2}}\text e^{\text i \omega x} $$ wich is normalised and found $$ \int |\psi'(x)|^2\;\text d x = \frac{1}{4\sigma^2}+\omega^2 $$ I guess in the most cases a term proportional to $1/\text{variance}$ will appear. And in the case of a simple phase another term will just be proportional to the frequency squared $\omega^2$.
The fact that is differentiable does not mean its derivative is square integrable, for that you need that $\psi\in H^1(\mathbb{R})\subset L^2(\mathbb{R})$, the Sobolev space. Hence you may have any order of magnitude (even infinite) for the $L^2$-norm of the derivative, given a generic differentiable $L^2$ function.