Proof: Distributing Divergence operator over cross product of two vectors

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Given:

$$F = (f_x, f_y, f_z)$$ $$G = (g_x, g_y, g_z)$$

Trying to prove that:

$$\nabla \cdot (F \times G) = G \cdot (\nabla \times F) - F \cdot (\nabla \times G)$$

So I start with:

$$\nabla \cdot (F \times G)$$

$$\nabla \cdot (F \times G) = \left( \frac{\partial}{\partial{x}},\ \ \ \frac{\partial}{\partial{y}},\ \ \ \frac{\partial}{\partial{z}} \right) \cdot \begin{vmatrix} \vec{\textbf{i}} & \vec{\textbf{j}} & \vec{\textbf{k}} \\ f_x & f_y & f_z \\ g_x & g_y & g_z \end{vmatrix}$$

$$\nabla \cdot (F \times G) = \left( \frac{\partial}{\partial{x}},\ \ \ \frac{\partial}{\partial{y}},\ \ \ \frac{\partial}{\partial{z}} \right) \cdot \vec{\textbf{i}} \begin{vmatrix} f_y & f_z \\ g_y & g_z \end{vmatrix} + \vec{\textbf{j}} \begin{vmatrix} f_x & f_z \\ g_x & g_z \end{vmatrix} + \vec{\textbf{k}} \begin{vmatrix} f_x & f_y \\ g_x & g_y \end{vmatrix}$$

$$\nabla \cdot (F \times G) = \left( \frac{\partial}{\partial{x}} \begin{vmatrix} f_y & f_z \\ g_y & g_z \end{vmatrix},\ \ \ \frac{\partial}{\partial{y}}\begin{vmatrix} f_x & f_z \\ g_x & g_z \end{vmatrix},\ \ \ \frac{\partial}{\partial{z}}\begin{vmatrix} f_x & f_y \\ g_x & g_y \end{vmatrix} \right)$$

$$\nabla \cdot (F \times G) = \frac{\partial}{\partial{x}}(f_y g_z - f_z g_y) - \frac{\partial}{\partial{y}}(f_x g_z - f_z g_x) + \frac{\partial}{\partial{z}}(f_x g_y - f_y g_x) $$

At this point, I get stuck in the proof...

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Hint: To get past this point, apply the product rule for partial derivatives. For example:

$$\frac{\partial}{\partial{x}}(f_y g_z) = \frac{\partial{f_y}}{\partial{x}} g_z + f_y \frac{\partial{g_y}}{\partial{x}}$$