Time derivative of dipole moment

While deriving the relation, $$\int_V \bar{J}d \tau = \frac{d\bar{p}}{dt}$$ in this question.

We have, $$\frac{d\bar{p}}{dt} = \frac{d}{dt}\int_V\rho \bar{r} d\tau \\ \implies \int _V \frac{\partial \rho}{\partial t}\bar{r}d\tau$$

My question is that shouldn't there be an additional term of $$+\int _V \rho\frac{\partial \bar{r}}{\partial t}d\tau$$ as a result of evaluating $$\int _V \frac{\partial}{\partial t}(\rho \bar{r} )d \tau$$

Also, can anyone elaborate as to how in one of the steps, $$\int_V \nabla .(r\bar{J})d\tau$$ is evaluated to be 0? I didn't quite understand the explanation in the first answer of the above mentioned link.

Also, here(eq 11), it's evaluated as $$\int_A (xJ)da$$ the value of which is stated to be 0, the reasoning of which I again fail to grasp.

It basically says that we can choose the boundary surface such that it lies outside all charges and currents and consequently J is zero on the surface. Couldn't we say the same thing while deriving the continuity equation by taking $$\oint_s J.da =0$$

I think I really have a messed up understanding of these concepts, so it would be really helpful if someone clarifies the doubts I have.