Can the del operator be "separated" like this? In class, we were asked to prove the identity $\nabla\times(\Omega\vec{V})=\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega$. One of the possible approaches involved separating the del operator into two operators,
\begin{equation}\nabla\equiv\nabla_{\Omega}+\nabla_{\vec{V}}\end{equation}
where $\nabla_{\Omega}$ differentiates $\Omega$ and leaves it operates on $\vec{V}$ zero, and vice versa. I know how to prove the identity, once I accept that $\nabla\equiv\nabla_{\Omega}+\nabla_{\vec{V}}$, but I find ti difficult to convince myself that it is true. Can del be separated like this?

The Del operator $\nabla$ operates on the independent coordinate variables, $x$, $y$, and $z$,not the dependent variables $\vec V(x,y,z)$ and $\Omega(x,y,z)$.  So, we have
$$\begin{align}
\nabla \times (\Omega \vec V)&=\sum_{i=1}^3 \hat x_i \frac{\partial}{\partial x_i}\times \left(\Omega \sum_{j=1}^3 \hat x_jV_j\right)\\\\
&=\sum_{i=1}^3 \sum_{j=1}^3 (\hat x_i \times \hat x_j )\left(\Omega \frac{\partial V_j}{\partial x_i}+V_j \frac{\partial \Omega}{\partial x_i}\right)\\\\
&=\Omega \nabla \times \vec V+\nabla \Omega \times \vec V\\\\
&=\Omega \nabla \times \vec V-\vec V\times \nabla \Omega 
\end{align}$$
as was to be shown!
