Classical Mechanics as an approximation of Quantum mechanics I want to show an equality:
We know from Ehrenfest's theorem that 
$$ \frac{d \langle x   \rangle(t)}{dt}=  \left\langle \frac{\partial H}{\partial p} \right\rangle
$$
I'd like to derive the following statement:
$$\frac{ d\langle x  \rangle(t)}{dt} = \frac{\partial  \langle H  \rangle}{\partial\langle p  \rangle}
$$ 
Because this would be in my Opinion the "total" classical limit: We observe a quantity energy and a quantity momentum, and that the derivation of energy with respect to momentum gives as the change of position with respect to time. 
Is it possible to derive the statement that I want to derive? If not exactly, what are the approximations that have to be made? I could show it under the condition that the hamilton operator and the momentum operator share eigenstates, but I'd like to know how mutch it is possible in general. 
Edit: 
By 
$$ 
 \frac{\partial  \langle H  \rangle}{\partial\langle p  \rangle}
$$ 
I mean that I look at the change of $\langle H \rangle$, when I change the state $\Psi$ in a way that changes the mean value of momentum. I could do that by constructing a function $f(p)$ that maps the number $p$ to a state $\Psi_p$ in a way that $\langle \Psi_p | \hat{p} | \Psi_p \rangle = p$ suffices. One example could be a gaussian wave package with a mean momentum p. Then I can take the derivative 
$
\frac{d}{dp} \langle \Psi_p | \hat{H} | \Psi_p \rangle
$. 
There is definitely not a unique way to define such a function $\Psi_p$, but if one can show the the equality I am looking for for one such function $\Psi_p$, without making use of the exact form that $\Psi_p$ takes, wouldn't the quantity $
\frac{\partial  \langle H  \rangle}{\partial\langle p  \rangle}
 = \frac{d}{dp} \langle \Psi_p | \hat{H} | \Psi_p \rangle
$ still be well defined?
Second edit: The origin of my question is the semiclassical model of electron dynamics, as it is explained in Chapter 12 of solid state physics by Ashcroft. Here the electron energy $\epsilon_k$ with $\hat{H} \Psi_k = \epsilon _k \Psi_k$ is effectively interpreted as hamiltonian, and wavepackets of electrons with mean crystal momentum $k$ are interpreted as classical particles. 
 A: First of all, you can't expect to recover general classical mechanics by simply making averages in quantum mechanics. Apart from very special cases, you can recover it only in the limit $\hslash\to 0$.
In such limit, something similar of what you expect can be proved. In particular, it holds when considering (squeezed) coherent states $C_{\hslash}(q,\xi)$ centred on $(q,\xi)\in\mathbb{R}^{2d}$ (with a suitable $\hslash$-dependence). First of all we suppose that $C_{\hslash}(q,\xi)$ is in the form domain of the Hamiltonian. Then
$$\lim_{\hslash\to 0}\langle H_{\hslash}\rangle_{C_{\hslash}(q,\xi)}=: h(q,\xi)$$
defines a function of the phase space. Such function has many nice properties of the classical energy, e.g.
$$h(t,q,\xi):=\lim_{\hslash\to 0}\langle H_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}= h(q,\xi)\; ;$$
i.e. it is an integral of motion.
On the other hand:
$$\lim_{\hslash\to 0}\langle x_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}=: q(t)\; ,\\
\lim_{\hslash\to 0}\langle p_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}=: \xi(t)\; ;$$
with $q(0)=q$ and $\xi(0)=\xi$.
Finally, it is possible to prove that $q(t)$ and $\xi(t)$ obey the expected equations of motion:
$$\left\{\begin{align}\dot{q}(t)&=\partial_{\xi}h(q(t),\xi(t))\\\dot{p}(t)&=-\partial_{q}h(q(t),\xi(t))\end{align}\right .\; .$$
