# Changing the partial derivative to the total derivative in Ampère-Maxwell's Law

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 $$\sum_i(\partial\vec{E}/\partial x_i)(dx_i/dt)$$?

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.