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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

If $f$, $g$ and $h$ are smooth vector fields, show that
$$\nabla\bullet(f(\nabla g\times\nabla h)) = \nabla f \bullet (\nabla
g \times \nabla h).$$


\textbf{Answer}

\begin{eqnarray*}
\nabla\bullet(f(\nabla g\times\nabla h)) & = & \nabla f \bullet
(\nabla g \times \nabla h)\\ & & + f\nabla\bullet(\nabla g \times \nabla
h)\\
& = & \nabla f \bullet (\nabla g \times \nabla h) +f((\nabla \times
\nabla g) \bullet \nabla h\\ & & - \nabla g \bullet (\nabla \times \nabla
h))\\
& = & \nabla f \bullet (\nabla g \times \nabla h) + \un{0} - \un{0}\\
& = & \nabla f \bullet (\nabla g \times \nabla h).
\end{eqnarray*}

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