Derivative of Vector Cross Product of Vector-Valued Functions/Proof 2

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Theorem

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 $\mathbf v = \mathbf a \times \mathbf b$.

Then:

\(\ds \dfrac {\d \mathbf v} {\d x}\) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf v} {x + h} - \map {\mathbf v} x} h\) Definition of Derivative of Vector-Valued Function
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \times \map {\mathbf b} {x + h} - \map {\mathbf a} x \times \map {\mathbf b} x} h\) Definition of $\mathbf v$
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \times \map {\mathbf b} {x + h} - \map {\mathbf a} {x + h} \times \map {\mathbf b} x + \map {\mathbf a} {x + h} \times \map {\mathbf b} x - \map {\mathbf a} x \times \map {\mathbf b} x} h\) rearranging
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \map {\mathbf a} {x + h} \times \dfrac {\map {\mathbf b} {x + h} - \map {\mathbf b} x} h + \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} - \map {\mathbf a} x} h \times \map {\mathbf b} x\) rearranging
\(\ds \) \(=\) \(\ds \mathbf a \times \dfrac {\d \mathbf b} {\d x} + \dfrac {\d \mathbf a} {\d x} \times \mathbf b\) Definition of Derivative of Vector-Valued Function

$\blacksquare$


Sources

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