Doubt on a vector calculus identity $\nabla\times(f\nabla g)$

Question

Let $f$ and $g$ be two scalar function of several real variables

$$ f,g:X\subseteq\mathbb{R}^3\rightarrow\mathbb{R} $$ $$ f,g\in C^2({\mathbb{R}^3}) $$ Calculate $$ \nabla\times(f\cdot\nabla{g}) $$ where $\nabla\times$ is the curl vector operator.

I've found the identity $$ \nabla\times(f\cdot\nabla{g}) = \nabla{f}\times\nabla{g} $$ where $\nabla{f}\times\nabla{g}$ is the vectorial product between the gradient respectively of function $f$ and $g$. Is it true?

Answer 1

Yes, it is true. Explicitly, both sides are

$$ (f_{{y}}g_{{z}}-f_{{z}}g_{{y}})\, \mathbf{i}+(-f_{{x}}g_{{z }}+f_{{z}}g_{{x}})\, \mathbf{j}+(f_{{x}}g_{{y}}-f_{{y}}g_{{ x}})\, \mathbf{k}$$

where subscripts denote partial derivatives.