How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$ Wikipedia indicates that the following relation is "easily shown": $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it.  I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$).  Can one of you walk me through it or link me to site that will?  Thanks.

Edit:
Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet):
$$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$
so
$$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$
where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$
and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$
Thus:
$$(x_iF(\vec p) - F(\vec p)x_i)\psi$$
$$=  x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$
$$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$
From here it doesn't look like those higher order terms are all going to cancel out.
 A: As @Qmechanic pointed out in a comment, we are free to use any operator representation. In momentum space, $\hat{\bf x} = + i \hbar \ \partial/\partial {\bf p} $ and $\hat{\bf p} = {\bf p}$, so
$$
\begin{eqnarray}
\left[\hat{x}_i,F\left(\hat{\bf p}\right)\right] &=& \left[i \hbar \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] \\
&=&  \frac{i \hbar}{f}\left[ \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] f \\
&=&  \frac{i \hbar}{f}\left( \frac{\partial}{\partial p_i}\left[F\left({\bf p}\right)f\right] - F\left({\bf p}\right)\frac{\partial f}{\partial p_i}\right)  \\
&=&  i \hbar \frac{\partial F\left({\bf p}\right)}{\partial p_i} \\
\end{eqnarray}
$$
A: Choose the momentum representation, 
$$x_i = i \hbar \frac{\partial}{\partial p_i}$$
distribute $i \hbar$ and act the commutator on vector $\psi$,
$$[x_i, F(\mathbf p)] \psi = i \hbar \left(\frac{\partial}{\partial p_i}(F(\mathbf{p}) \space \psi) -F(\mathbf p) \frac{\partial }{\partial p_i} \psi \right)$$ 
and apply the product rule:
$$= i \hbar \left(\frac{\partial F(\mathbf p )}{\partial p_i} \psi + F(\mathbf p) \frac{\partial \psi}{\partial p_i}  - F(\mathbf p) \frac{\partial \psi}{\partial p_i} \right)$$
$$= i \hbar\frac{\partial F(\mathbf p )}{\partial p_i} \psi.$$
We left $\psi$ unspecified, so:
$$ [ x_i, F(\mathbf p ) ] = i \hbar \frac{\partial F( \mathbf p )}{\partial p_i}$$
A: The commutation of two variables, in some cases, can be related to Poisson Bracket via
$$
\left[\hat A,\,\hat B\right]=i\hbar\left\{\hat A,\,\hat B\right\}
$$
Thus,
$$
\left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)\tag{1}
$$
Formally $\hat A=A(\hat q,\,\hat p)$ and $\hat B= B(\hat q,\,\hat p)$. You should be able to use (1) to solve your problem. Note, though, that this is not a solution that works in all cases (cf., this and this that Qmechanic pointed out) as it is an approximation that only holds under certain cases.
A general solution involves the Moyal bracket,
$$
\left[\hat A,\,\hat B\right]\sim\{\{\hat A\,,\hat B\}\}\sim \hat A\star \hat B-\hat B\star\hat A
$$
where $\star$ denotes the Moyal star-product (see answers on either this post or this post for more on the Moyal product). The above then can be written as
$$
\{\{\hat A,\,\hat B\}\}=\{\hat A,\,\hat B\}+\mathcal{O}(\hbar^2)
$$
where $\{\cdot,\,\cdot\}$ here is the above Poisson bracket and $\mathcal{O}(\hbar^2)$ are corrections (referred to as deformations if the Poisson bracket).
Thus, Equation (1) becomes
$$
\left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)+\mathcal{O}(\hbar^2)\tag{2}
$$
