My question is the following: do there exist Borel measures $\mu$ and $\nu$ such that, given a differentiable function $f$, we have $$ f'(0) = \int_\mathbb{R} f d\mu - \int_\mathbb{R} f d\nu? $$
I've been thinking about the one-to-one correspondence between Borel measures and (left) continuous increasing functions $F$ given by $$ \mu((a,b]) := F(b) - F(a). $$ I figure if I can choose an appropriate $F$, then the result would follow. I was also thinking that maybe the Riesz representation theorem is another approach, since differentiation is a linear map. I'd appreciate any help.
This is impossible. The idea is that a function which is compactly supported and uniformly close to 0 must have small integral, but its derivative at 0 can still be big.
Let $\phi$ be some nonnegative smooth cutoff function which has support inside $[-1,1]$ and equals 1 on some neighborhood of 0. If your equation is to hold for $\phi$, then in particular $\int \phi\,d\mu$ and $\int \phi\,d\nu$ must both be finite. Now consider the function $f_n(t) = \frac{1}{n} \phi(t) \sin(n^2 t)$. Note that $|f_n| \le \phi$ and $f_n \to 0$ pointwise (even uniformly), so $\int f_n\,d\mu \to 0$ and $\int f_n\,d\mu \to 0$ by dominated convergence. Thus the right side of the proposed equation goes to $0$. But $f_n'(0) = n \to \infty$.