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Ampère-Maxwell's Law, in it's integral form, is $$ \oint_{\partial \Sigma}\vec{B}\cdot d \vec{l}=\mu_0\left(\iint_{\Sigma}\vec{J}\cdot d \vec{A}+\epsilon_0\frac{d }{d t}\iint_{\Sigma}\vec{E}\cdot d\vec{A}\right), $$ let's take $\vec{J}=0$ to make things easier. We can, then, write it as $$\iint_{\Sigma}\vec{\nabla}\times\vec{B}\cdot d \vec{l}=\mu_0\epsilon_0\frac{d }{d t}\iint_{\Sigma}\vec{E}\cdot d\vec{A},$$

and since $\Sigma$ is arbitrary, we get

$$\vec{\nabla}\times\vec{B}=\mu_0\epsilon_o\frac{\partial}{\partial t}\vec{E}.$$

Here's my question: is there a quick way to know if can we write $\partial/\partial t$ as $d/dt$ besides evaluating $\sum_i(\partial\vec{E}/\partial x_i)(dx_i/dt)$?

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It;s a variant of the notation in Leibnitz' formula which says $$ \frac d{dt} \int^{b(t)}_{a(t)} f(x,t) dx= \frac{db}{dt} f(b(t))- \frac{da}{dt}f(a(t)) + \int_{a(t)}^{b(t)} \frac{\partial f }{\partial t}(x,t)dx. $$ On the LHS the expression only depends on $t$ so there is no need for a partial derivative. But in the integral on the RHS the expression $f(x,t) $ depends of both $x$ and $t$ so you need to use the partial derivative symbol for the time derivative because $x$ is being understood as being fixed.

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