## Identity for a 3 - form on a Vector Function in R3

Theorem
Let
$\omega^3 =g(\mathbf{x}) dx_1 \wedge dx_2 \wedge dx_3$
be a 3 form in
$\mathbb{R}^3$
and let
$\mathbf{f} (\mathbf{x}) : \mathbb{R}^3 \rightarrow \mathbb{R}^3$
be a differentiable function.
Then
$\omega^3_{\mathbf{f}(\mathbf{u})} (\frac{\mathbf{f}}{\partial u_1 } , \frac{\mathbf{f}}{\partial u_2 }, \frac{\mathbf{f}}{\partial u_3 }) =g(\mathbf{f} (\mathbf{u}) (\frac{f_1}{\partial u_1 } , \frac{f_2}{\partial u_2 }, \frac{f_3}{\partial u_3 })$

Proof
\begin{aligned} \omega^3_{\mathbf{f}(\mathbf{u})} (\frac{\mathbf{f}}{\partial u_1 } , \frac{\mathbf{f}}{\partial u_2 }, \frac{\mathbf{f}}{\partial u_3 }) &= g (\mathbf{f} (\mathbf{u})) dx_1 \wedge dx_2 \wedge dx_3 (\frac{\mathbf{f}}{\partial u_1 } , \frac{\mathbf{f}}{\partial u_2 }, \frac{\mathbf{f}}{\partial u_3 }) \\ &= g (\mathbf{f} (\mathbf{u})) \left| \begin{array}{ccc} \frac{\partial f_1}{\partial u_1} & \frac{\partial f_1}{\partial u_2} & \frac{\partial f_1}{\partial u_3} \\ \frac{\partial f_2}{\partial u_1} & \frac{\partial f_2}{\partial u_2} & \frac{\partial f_2}{\partial u_3} \\ \frac{\partial f_3}{\partial u_1} & \frac{\partial f_3}{\partial u_2} & \frac{\partial f_3}{\partial u_3} \end{array} \right| \\ &= g(\mathbf{f} (\mathbf{u})) \frac{\partial (f_1 , f_2 , f_3)}{\partial (u_1 ,u_2 , u_3)} \end{aligned}