Consider a scalar potential $\phi$, a vector potential $\mathbf{A}$, an electric field satisfying $\mathbf{E}=-\mathbf{\nabla}\phi-\dfrac{\partial}{\partial t}\mathbf{A}$, and a magnetic field satisfying $\mathbf{B}=\mathbf{\nabla}\times\mathbf{A}$.
I need to prove that $\mathbf{E}$ and $\mathbf{B}$ satisfy Maxwell's first equation in the differential form $$\nabla\times\mathbf{E}=-\partial_t \mathbf{B}$$
So I started off just by rearranging to obtain that:
$\nabla\times\mathbf{E}+\partial_t \mathbf{B}=0$
But
$\mathbf{\nabla} \times \mathbf{E} + \partial_t \mathbf{B}$
$=\mathbf{\nabla}\times\left(-\mathbf{\nabla}\phi-\dfrac{\partial}{\partial t}\mathbf{A}\right)+\partial_t \mathbf{B} $
$=-\mathbf{\nabla}(\mathbf{\nabla}\phi)-\mathbf{\nabla}\partial_t\mathbf{A} + \partial_t\mathbf{B}$
But I'm unsure as of what to do next...and if what I've done is actually correct? Thanks for any help
Your last line isn't quite correct.
You didn't distribute the curl correctly: it should be $\nabla\times\nabla\phi$ and $\nabla\times\partial_t\mathbf{A}$. But the curl of a gradient is zero, and for $\nabla\times \partial_t\mathbf{A}$, you can exchange the operators, because $\nabla$ only encodes derivatives with respect to spatial coordinates, whereas $\partial_t$ only deals with time. So you get $$-\underbrace{\nabla\times\nabla\phi}_{0}-\partial_t\nabla\times\mathbf{A}+\partial_t\mathbf{B}=-\partial_t\mathbf{B}+\partial_t\mathbf{B}=0$$ where I've used the definition of the vector potential: $\nabla\times\mathbf{A}=\mathbf{B}$.