If
\[f, \;\]
are differentiable functions then \[(\mathbf{\nabla}f) \times (\mathbf{\nabla}g)\]
is solenoidal, so that \[\mathbf{\nabla} \cdot(\mathbf{\nabla}f) \times (\mathbf{\nabla}f)=0\]
(1).
ProofUse the identity
\[\mathbf{\nabla} \cdot (\mathbf{a} \times \mathbf{b})=\mathbf{b} \cdot (\mathbf{\nabla} \times \mathbf{a})- \mathbf{a} \cdot (\mathbf{\nabla} \times \mathbf{b}) \]
so that (1) becomes\[\mathbf{\nabla} \cdot ((\mathbf{\nabla} f \times \mathbf{\nabla} g) =\mathbf{\nabla} g \cdot (\mathbf{\nabla} \times \mathbf{\nabla} f) - \mathbf{\nabla} f \cdot (\mathbf{\nabla} \times \mathbf{\nabla} g)\]
But
\[\mathbf{\nabla} \times \mathbf{\nabla} f = \mathbf{\nabla} \times \mathbf{\nabla} g =0\]
Hence
\[\mathbf{\nabla} \cdot(\mathbf{\nabla}f) \times (\mathbf{\nabla}f)=0\]