Time derivative of curl I'm confused about the time derivative of the curl. If the following true? Where do I go wrong?
$$\frac { \partial } { \partial t} \ \vec{\nabla} \times \vec{v} = \frac{\partial \vec \nabla }{ \partial t} \times \vec v + \vec \nabla \times \frac { \partial \vec v }{\partial t} $$
The reason I'm confused is that the second term looks exactly like the first term, but the second term doesn't necessarily/immediately look like it vanishes. 
If it does vanish, is it because $$ \frac {\partial \vec v}{\partial t} = \vec a = \vec \nabla \Phi $$ and since $\mathrm {curl(grad (\Phi))} = 0$, my thought that the first term is just like the LHS is true? 
Something seems wrong here to me because it seems too strong a hypothesis to consider the time derivative of the vector field, $\vec v$ conservative.
 A: It's best when you're working with not-true-vectors such as curl to calculate from first principles: $\frac{\partial}{\partial t}\vec{\nabla}\times\vec{v}$ has $i$th component $\epsilon_{ijk}\frac{\partial}{\partial t}\frac{\partial}{\partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $\epsilon_{ijk}\frac{\partial}{\partial x_j} \frac{\partial}{\partial t}v_k$, which is the $i$th component of $\vec{\nabla}\times\frac{\partial}{\partial t}\vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $\frac{\partial}{\partial t}\vec{\nabla}=\vec{0}$, whatever that would mean.
A: The notation $\nabla \times \,\cdot\,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $\nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${\bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $\Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$\frac{d}{dt}(\nabla \times {\bf v})$$ is
$$\frac{\partial}{\partial t}\left(\frac{\partial v^2}{\partial z} - \frac{\partial v^3}{\partial y}\right) = \frac{\partial}{\partial z}\left(\frac{\partial v^2}{\partial t}\right) - \frac{\partial}{\partial y}\left(\frac{\partial v^3}{\partial t}\right) .$$
A: What you're simply doing is taking the derivative of the curl, which happens to be a time derivative. Say you have a $C^2$ scalar function $A=A(x,y,z,t)$. Assuming you're working in Cartesian coordinates, then all you have to do is go through the motion and compute!
$$
 \begin{align}
  \vec\nabla\times\frac{\partial \vec A}{\partial t}&=
  \begin{vmatrix}
   \hat x & \hat y & \hat z\\
   \partial_x & \partial_y &\partial_z\\
   \partial_t A_1 & \partial_t A_2 &\partial_t A_3
  \end{vmatrix}\\[10pt]
  &=\bigg[\frac{\partial^2 A_3}{\partial y\partial t}-\frac{\partial ^2 A_2}{\partial z\partial t}\bigg]\hat x
   -\bigg[\frac{\partial ^2A_3}{\partial x\partial t}-\frac{\partial ^2A_1}{\partial z\partial t}\bigg]\hat y
   +\bigg[\frac{\partial ^2A_2}{\partial x\partial t}-\frac{\partial ^2A_1}{\partial y\partial t}\bigg]\hat z\\[10pt]
  &=\frac{\partial}{\partial t}\bigg[\frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z}\bigg]\hat x
   -\frac{\partial}{\partial t}\bigg[\frac{\partial A_3}{\partial x}-\frac{\partial A_1}{\partial z}\bigg]\hat y
   +\frac{\partial}{\partial t}\bigg[\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}\bigg]\hat z\\[10pt]
  &=\frac{\partial}{\partial t}(\vec\nabla \times \vec A)\;\blacksquare
 \end{align}
$$
This proof works both ways! Note that A being $C^2$ is sufficient enough in order to use the mixed-partials equivalence theorem.
