Ampères law: Getting $\nabla \times \vec H = \vec J_{free} + \frac{\partial \vec D}{\partial t}$ by taking the cross product I've seen two different versions of Ampère's law and I'm having trouble 
connecting them:
$$\nabla \times \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t}$$
$$\nabla \times \vec H = \vec J_{free} + \frac{\partial \vec D}{\partial t}$$

Trying to "derive" the second version of it
Assumption: $\vec H = \frac{\vec B}{\mu_0} - \vec M$
$ \nabla \times \vec H = \nabla \times (\frac{\vec B}{\mu_0} - \vec M) = \vec J +  \epsilon_0 \frac{\partial \vec E}{\partial t} - \vec J_m = \vec J_{free} + \epsilon_0 \frac{\partial \vec E}{\partial t} \stackrel{?}{\neq}  \vec J_{free} + \frac{\partial \vec D}{\partial t} $
Since according to my understanding $\vec D = \epsilon \vec E \neq \epsilon_0 \vec E$
Where am I going wrong?
 A: The reason for this discrepancy is that your first equation is, in fact, incorrect. In a dielectric media, the field $\boldsymbol{E}$ is not the only one to affect the flux, but also $\boldsymbol{P}$. The latter give rise to bound-charge current density
$$\boldsymbol{J}_{\rm bound}=\mu_{0}\frac{\partial\boldsymbol{P}}{\partial t}$$
This contribution can be combined with $\mu_{0}\varepsilon_{0}\frac{\partial\boldsymbol{E}}{\partial t}$ to yield $\mu_{0}\frac{\partial\boldsymbol{D}}{\partial t}$, giving the correct term that you are missing.

Reference:
[1] Purcell, E. M., Morin, D. J., Electricity and Magnetism, 3rd Edition, pp. 505-507.
A: Igonring the non-local electromagnetic response (quadrupole and higher effects) one has
$\mathbf{D}=\epsilon_0 \mathbf{E}+\mathbf{P}$, where $\mathbf{P}$ is the polarization = density of electric dipoles per volume.
Also one has $\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M} $, where $\mathbf{M}$ is the magnetization = density of magnetic dipoles per volume.
The full current density is: $\mathbf{J}=\mathbf{J}_{free}+\frac{\partial\mathbf{P}}{\partial t}+\mathbf{\nabla}\times\mathbf{M}$
This shoud be sufficient to proove everything.Start with $\mathbf{\nabla}\times\mathbf{H}=\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$
The tricky bit is to show how the magnetization/polarization relate to current density. One can do it by considering point-like electric and magnetic dipoles. I don't have time to show it here (ask me if you are actually interested in it)
