## Proof That a Cross Product of Gradients is Solenoidal

Theorem
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). Proof
Use 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
$\mathbf{\nabla} \cdot(\mathbf{\nabla}f) \times (\mathbf{\nabla}f)=0$