Derivative of Dot Product of Vector-Valued Functions/Proof 1
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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:
- $\mathbf a: x \mapsto \tuple {\map {a_1} x, \map {a_2} x, \ldots, \map {a_n} x}$
- $\mathbf b: x \mapsto \tuple {\map {b_1} x, \map {b_2} x, \ldots, \map {b_n} x}$
Then:
\(\ds \map {\frac \d {\d x} } {\mathbf a \cdot \mathbf b}\) | \(=\) | \(\ds \map {\frac \d {\d x} } {\sum_{i \mathop = 1}^n a_i b_i}\) | Definition of Dot Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i b_i}\) | Sum Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {\map {\frac \d {\d x} } {a_i} b_i + a_i \map {\frac \d {\d x} } {b_i} }\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i} b_i + \sum_{i \mathop = 1}^n a_i \map {\frac \d {\d x} } {b_i}\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}\) | Definition of Dot Product |
$\blacksquare$
Sources
- 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 7$
- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): $\S 12.2$