Proof Using Faraday's Law and Ampere's Law

Question

For $E = E(x, y,z, t)$ and $B = B(x, y, z, t)$,

$\nabla \times E = -\frac{\delta B}{\delta t}$ and $\nabla \times B = \frac{1}{c^2} \frac{\delta E}{\delta t}$,

how can I show that $\nabla \times (\nabla \times E) = -\frac{1}{c^2} \frac{\delta^2 E}{\delta t^2}$?

If someone could point me in the right direction, it would be greatly appreciated.

Answer 1

Observe \begin{align} \nabla \times (\nabla \times E) = \nabla\times\left(-\frac{\delta B}{\delta t} \right) = -\nabla \times \frac{\delta B}{\delta t} = -\frac{\delta}{\delta t}\left(\nabla\times B \right) = -\frac{1}{c^2}\frac{\delta^2 E}{\delta t^2}. \end{align}

Of course you still need to justify why \begin{align} -\nabla \times \frac{\delta B}{\delta t} = -\frac{\delta}{\delta t}\left(\nabla\times B \right), \end{align} i.e. you need to show why you can drag the time derivative out of the cross product.