I'm pretty new to tensor calculus; I would like to prove, as an exercise, the following vector calculus identity:
$$ \partial_t \left( \nabla \times \vec{F} \right) = \nabla \times \partial_t \vec{F} $$
using tensors and tensor notation. I've started by setting up a tensor-like expression for the $i$-th component of the curl of the general vector $\vec{F}$, thus:
$$ \left( \nabla \times \vec{F} \right)^i = \varepsilon^{ijk} \nabla_j F_k $$
Now I take the time derivative of the expression:
$$ \partial_t \left( \nabla \times \vec{F} \right)^i = \partial_t \left( \varepsilon^{ijk} \nabla_j F_k \right) $$
At this point I started having doubts about the procedure. Following my reasoning I should apply the product rule and hope that the only non-zero piece of that expression will be $\varepsilon^{ijk} \nabla_j \left( \partial_t F_k \right) = (\nabla \times \partial_t \vec{F})^i$. I'm not even sure about the $\partial_t F_k$ on the l.h.s.: should that be considered a twice covariant tensor? Anyway applying the product rule:
$$ \partial_t \left( \varepsilon^{ijk} \nabla_j F_k \right) = \left( \partial_t \varepsilon^{ijk} \right) \nabla_j F_k + \varepsilon^{ijk} \left( \partial_t \nabla_j \right) F_k + \varepsilon^{ijk} \nabla_j \left( \partial_t F_k \right) $$
Now the term containing $\partial_t \varepsilon^{ijk}$ vanishes. It should also vanish the term with $\partial_t \nabla_j$ in it, but I'm not sure how to justify it. Therefore I would remain with:
$$ \varepsilon^{ijk} \nabla_j \left( \partial_t F_k \right) = \left( \nabla \times \partial_t \vec{F} \right) $$
Is this reasoning correct? If yes, how would you justify the fact that $\varepsilon^{ijk} \left( \partial_t \nabla_j \right) F_k$ vanish? And do I have to consider $\partial_t F_k$ as a twice covariant tensor $T_{tk}$ or should I treat $\partial_t$ like an intrinsic derivative $\delta_t$?
P.S. Thanks in advance for your answers and excuse my poor Engliush, I'm still practising it!