Is there any good reason/argument/heuristic why one can approximate forces by approximating the potentials? My question
Is there any good reason/argument/heuristic why one can approximate forces by approximating the potentials? (As a concrete example, in Electrostatics.)
Motivation for the question
I am studying for a (non quantum) electrodynamics course (loosely following Landau and Lifshitz). In electrostatics, we assume that the charge distribution is contained in some bounded part around the origin; and we want to calculate (or rather: approximate) the field  in a large distance $r$.
This is done the following way:
Approximate the potential $\phi(r \vec n)=\frac{Q}{r}+\frac{n_i d_i}{r^2}+O(\frac{1}{r^3})$ (here, $Q$ is the total charge and $\vec d$ the dipole moment). So obviously if we stop at the $\frac{1}{r^2}$ term, the resulting $\phi'$ is a "good" approximation to the potential $\phi$. Then it seems to be implicitly assumed that the field $\vec E'$ given by $\phi'$ is a "good" approximation to the "true" $\vec E$ given by $\phi$.
This sounds physically rather convincing and natural. But obviously, it is not generally true that $f\approx g$ implies $f'\approx g'$.
Some musings
Maybe the answer is that the forces are not that important anyway:
Many properties of the systems can be described "in a continuous way" by the potential. In Electrostatics this is rather obvious; in more general electrodynamics or in settings where there is no clearly preserved total Energy this becomes more and more fuzzy to me. 
Btw., I am aware that there is a very general answer along the lines of: "in this family of PDEs, the solutiuon depends continuously from the input, so of course a good approximation to the potential will result in very accurate equations of motion". However, this is not terribly satisfying, since the point here seems to be to estimate (at least in some fuzzy way) how good our approximations is. (For example, this general answer would not tell me why in the example above it is reasonable to ignore $O(\frac1{r^3})$ to get the "first order approximation" to some equation of motion; the $O(\frac1{r^3})$ might have more influence than the $\frac1{r^2}$ part).
 A: Suppose you have any analytical function $f(x)$ and you do an Taylor expansion:
$$
f_T(x) = \mathcal{T}_\infty[f](x)= \sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n
$$
For analytical functions $f_T = f$. Then the derivative of $f_T(x)$ is:
$$
f_T'(x) = \sum\limits_{n=1}^\infty \frac{f^{(n)}(0)}{(n-1)!}x^{n-1}
= \sum\limits_{n=0}^\infty \frac{f^{(n+1)}(0)}{n!}x^{n}
= \mathcal{T}_\infty[f'](x)
$$
and since the derivative of an analytical function is again analytic:
$$
f_T'(x) = f'(x)
$$
So there is a general relation between the Taylor expansion of a function and its derivative.
Now suppose you stop the expansion at order $d$. Then:
$$
f_T(x) = \mathcal{T}_d[f](x) = \sum\limits_{n=0}^d \frac{f^{(n)}(0)}{n!}x^n + \mathcal{O}(x^{d+1})
$$
Again the derivative gives:
$$
f_T'(x) = \sum\limits_{n=1}^d \frac{f^{(n)}(0)}{(n-1)!}x^{n-1} + \mathcal{O}(x^d) = 
\sum\limits_{n=0}^{d-1} \frac{f^{(n+1)}(0)}{n!}x^{n} + \mathcal{O}(x^d)
= \mathcal{T}_{d-1}[f'](x) + \mathcal{O}(x^d)
$$
This means that differentiating the expansion to order $d$ yields the derivative of the original function approximated to order $d-1$.
The same reasoning can be used to show that approximating the potential to order $\frac{1}{r^2}$ yields the field approximated to order $\frac{1}{r^3}$. For this let $y = \frac{1}{r}$
$$ \phi'(r) = \frac{\partial\phi}{\partial y}\frac{\partial y}{\partial r} = -\frac{1}{r^2} \phi'(y) $$
The only difference here is that there is an additional factor of $\frac{1}{r^2}$ which means that an approximation of of $\phi$ in $d$-th order of $y = \frac{1}{r}$ results in a $d+1$-th order approximation of $\phi'$.
Therefore the quality of the approximation in your example is very well defined. If the potential is approximated to order $\frac{1}{r^2}$, then the field is approximated to order $\frac{1}{r^3}$.
The same probably extends to more than one dimension, although I haven't checked that.
Also note that in terms of the general quality of Taylor expansions there are ways to estimate the remainder term or give bounds on the remainder term, see for one example Wikipedia. If this term is calculated the minimum distance for the approximation to be good can be calculated. However the general statement of such approximations is, that for some (not necessarily known) minimal distance the Taylor expansion gives a good result and an exact one in infinite distance.
EDIT: Due to comment discussions I want to add:
Every potential has to satisfy the Poisson equation:
$$\nabla^2 \phi(\vec r) = \rho(\vec r)$$
where $\rho$ is the charge distribution associated with the potential. Now suppose the charge distribution is contained in a ball of radius $R$ from the origin. Then outside this box the Poisson equation reduces to the Laplace equation. Any solution for the potential thus has to satisfy the Laplace equation for $r > R$. The real solutions of the Laplace equation are harmonic functions, which are always real-analytical. For $r > R$ therefore the potential is analytical and can be approximated with a Taylor series around $r = \infty$. Then for the field the first part of my answer applies.
