= Solution
In <Einstein notation>,
$$
\begin{aligned}
[\nabla\times(\psi A)]_i
&=\epsilon_{ijk}\partial_j(\psi A_k)\\
&=\psi\epsilon_{ijk}\partial_jA_k
+\epsilon_{ijk}(\partial_j\psi)A_k.
\end{aligned}
$$
This is
$$
\boxed{\nabla\times(\psi A)
=\psi\nabla\times A+\nabla\psi\times A}.
$$
Similarly,
$$
\begin{aligned}
\nabla\cdot(A\times B)
&=\partial_i(\epsilon_{ijk}A_jB_k)\\
&=\epsilon_{ijk}(\partial_iA_j)B_k
+\epsilon_{ijk}A_j(\partial_iB_k)\\
&=B\cdot(\nabla\times A)-A\cdot(\nabla\times B),
\end{aligned}
$$
after cyclically relabelling the first term and swapping two indices in the second. Hence
$$
\boxed{\nabla\cdot(A\times B)
=B\cdot\nabla\times A-A\cdot\nabla\times B}.
$$
Back to article page
