Need help to prove the following equation [closed]

Question

I am able to show the equation at (2) but I am not sure how to show the equation at (3). Any help is greatly appreciated

Answer 1

$\displaystyle \nabla \times \frac{dE}{dt}$ is equivalent to $\displaystyle \frac{d}{dt} \left( \nabla \times E \right)$ This can be shown by breaking down the curl. For example, $\displaystyle \frac{d}{dx} \frac{d}{dt} E = \frac{d}{dt} \frac{d}{dx} E$ since $x$ is independent of $t$ (Newtonian speeds)

Take the curl of one of the existing equations (same concept works for E & H): $\displaystyle \nabla \times (\nabla \times E) = \mu_0 \frac{d}{dt} \left( \nabla \times H \right)$

Now use the fact that $\displaystyle \nabla \times H = \epsilon_0 \frac{dE}{dt}$

$\displaystyle \nabla \times (\nabla \times E) = \mu_0 \frac{d}{dt} \left( \epsilon_0 \frac{dE}{dt} \right)$