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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

Given that $\phi$ and $\psi$ are smooth scalar fields, show that
$$\nabla\times(\phi\nabla\psi) = - \nabla\times(\psi\nabla\phi) =
\nabla\phi \times \nabla\psi.$$


\textbf{Answer}

\begin{eqnarray*}
\nabla\times(\phi\nabla\psi) & = & \nabla\phi \times \nabla\psi +
\phi\nabla \times \nabla\psi\\
& = & \nabla\phi \times \nabla\psi\\
- \nabla\times(\psi\nabla\phi) & = & -\nabla \psi \times \nabla \phi -
\psi \nabla \times \nabla\phi\\
& = & \nabla\phi \times \nabla\psi.
\end{eqnarray*}

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