Partial derivative of a vector

Question

I'm trying to show: $\displaystyle \frac{\partial} {\partial t}( \nabla(\phi))= \nabla\frac{\partial \phi} {\partial t} $

Am I allowed to write:

$\displaystyle \frac{\partial} {\partial t}\begin{bmatrix} \frac{\partial\phi} {\partial x}\\\ \frac{\partial\phi} {\partial y}\\\ \frac{\partial\phi} {\partial z} \end{bmatrix}=\displaystyle \begin{bmatrix} \frac{\partial^2\phi} {\partial t\partial x}\\\ \frac{\partial^2\phi} {\partial t\partial y}\\\ \frac{\partial^2\phi} {\partial t\partial z} \end{bmatrix}$

Answer 1

I think it is correct. $$ \frac{\partial}{\partial t} \nabla \phi = \partial_t \sum_i e_i \partial_i \phi =^{(*)} \sum_i e_i \partial_t \partial_i \phi $$ writing $\partial_t = \partial / \partial t$, $\partial_i = \partial / \partial x_i$, for some basis vectors $e_i$.

Your second equation corresponds to $(*)$.

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It continues with $$\sum_i e_i \partial_t \partial_i \phi = \sum_i e_i \partial_i \partial_t \phi = \nabla\frac{\partial}{\partial t} \phi$$ if $[\partial_t, \partial_i] = \partial_t \partial_i - \partial_i \partial_t=0$ for $\phi$. And this link is useful.

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