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

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?

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