Derivative of Dot Product of Vector-Valued Functions/Proof 2
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Theorem
Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
The derivative of their dot product is given by:
- $\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$
Proof
Let $\mathbf v = \mathbf a \cdot \mathbf b$.
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \map {\mathbf a} x \cdot \map {\mathbf b} x} h\) | Definition of $\mathbf v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \map {\mathbf a} {x + h} \cdot \map {\mathbf b} x + \map {\mathbf a} {x + h} \cdot \map {\mathbf b} x - \map {\mathbf a} x \cdot \map {\mathbf b} x} h\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \map {\mathbf a} {x + h} \cdot \dfrac {\map {\mathbf b} {x + h} - \map {\mathbf b} x} h + \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} - \map {\mathbf a} x} h \cdot \map {\mathbf b} x\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf a \cdot \dfrac {\d \mathbf b} {\d x} + \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b\) | Definition of Derivative of Vector-Valued Function |
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products: $(3.3)$