Let $\psi\in C_c^{\infty}(\mathbb{R})$, with support away from zero and let $t\mapsto \phi(t)$ be a function of time satisfying the equality:
$$\phi(t)=\int_0^t\psi(s)\,ds$$
and therefore:
$$\psi(x)=\frac{d}{dx}\phi(x).$$
Suppose that we begin with an analogue sampled $\phi[n]:=\phi(n\Delta t)$, where $\Delta t$ is our sampling interval, and we want to use this to reconstruct $\psi$. Then we can use the following $\operatorname{sinc}$ based interpolation formula to first reconstruct
$$\phi(t)=\sum_{n=-\infty}^{\infty}\phi[n]\operatorname{sinc}\left(\frac{t-n\Delta t}{\Delta t}\right).$$
Consequently, we can reconstruct $\psi(x)$ from $\phi[n]$ using
$$\psi(x)=\sum_{n=-\infty}^{\infty}\phi[n]\frac{d}{dx}\operatorname{sinc}\left(\frac{x-n\Delta t}{\Delta t}\right).$$
However, I am really struggling to account for the derivative of $\operatorname{sinc}$ in my Matlab code.
I can, for instance, use:
function psi = interpolate(ts,phin,x)
psi = 0;
for n=1:length(ts)
psi = x+phin(n)*diff(sinc((x-n.*T)./T));
end
psi = [psi 0];
But the problem is that this is a numerical approximation of the derivative and my plot is essentially shifted due to the vector I remove to allow me to plot $x(t)$ with respect to $t$.
Red = The actual function; Blue = The reconstructed function
Of course, I can analytically deduce the actual derivative of $\operatorname{sinc}$, but as Fabian noted in his answer, the sampled version doesn't resemble the actual derivative of $\operatorname{sinc}$ at all.
The sinc function is heavily oscillation. Does it is not a good idea to take its derivative; or in other words, the derivative of $\phi(x)$ via the sampled expression will not have a lot to do with the "real" derivative.
In any case, it would be more useful and stable to rely on standard discrete approximations to $\frac{d}{dx} \phi(x)$ such as the forward, backward, or central difference.