Theorem: Derivative of Vector Cross Product of Vector-Valued Functions

Let $\mathbf a: \R \to \R^3$ and $\mathbf b: \R \to \R^3$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their vector cross product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b} = \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mathbf b} {\d x}$

Proof:
Let:
:$\mathbf a: x \mapsto \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$
:$\mathbf b: x \mapsto \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b
      | r = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}
}}
{{eqn | r = \begin {bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end {bmatrix}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b}
      | r = \begin {bmatrix} \map {\dfrac \d {\d x} } {a_2 b_3 - a_3 b_2} \\ \map {\dfrac \d {\d x} } {a_3 b_1 - a_1 b_3} \\ \map {\dfrac \d {\d x} } {a_1 b_2 - a_2 b_1} \end {bmatrix} 
      | c = Differentiation of Vector-Valued Function Componentwise
}}
{{eqn | r = \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 + a_2 \dfrac {\d b_3} {\d x} - \dfrac {\d a_3} {\d x} b_2 - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 + a_3 \dfrac {\d b_1} {\d x} - \dfrac {\d a_1} {\d x} b_3 - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 + a_1 \dfrac {\d b_2} {\d x} - \dfrac {\d a_2} {\d x} b_1 - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}
      | c = Product Rule for Derivatives, Linear Combination of Derivatives
}}
{{eqn | l = \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mathbf b} {\d x}
      | r = \begin {bmatrix} \dfrac {\d a_1} {\d x} \\ \dfrac {\d a_2} {\d x} \\ \dfrac {\d a_3} {\d x} \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix} + \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} \dfrac {\d b_1} {\d x} \\ \dfrac {\d b_2} {\d x} \\ \dfrac {\d b_3} {\d x} \end{bmatrix}
}}
{{eqn | r = \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 - \dfrac {\d a_3} {\d x} b_2 \\ \dfrac {\d a_3} {\d x} b_1 - \dfrac {\d a_1} {\d x} b_3 \\ \dfrac {\d a_1} {\d x} b_2 - \dfrac {\d a_2} {\d x} b_1 \end {bmatrix} + \begin {bmatrix} a_2 \dfrac {\d b_3} {\d x} - a_3 \dfrac {\d b_2} {\d x} \\ a_3 \dfrac {\d b_1} {\d x} - a_1 \dfrac {\d b_3} {\d x} \\ a_1 \dfrac {\d b_2} {\d x} - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 - \dfrac {\d a_3} {\d x} b_2 + a_2 \dfrac {\d b_3} {\d x} - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 - \dfrac {\d a_1} {\d x} b_3 + a_3 \dfrac {\d b_1} {\d x} - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 - \dfrac {\d a_2} {\d x} b_1 + a_1 \dfrac {\d b_2} {\d x} - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}
}}
{{eqn | r = \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 + a_2 \dfrac {\d b_3} {\d x} - \dfrac {\d a_3} {\d x} b_2 - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 + a_3 \dfrac {\d b_1} {\d x} - \dfrac {\d a_1} {\d x} b_3 - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 + a_1 \dfrac {\d b_2} {\d x} - \dfrac {\d a_2} {\d x} b_1 - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b}
}}
{{end-eqn}}
{{qed}}