Derivative of Dot Product of Vector-Valued Functions/Proof 1

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:

$\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$

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 $

$=$

$\ds \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i b_i}$

$\ds $

$=$

$\ds \sum_{i \mathop = 1}^n \paren {\map {\frac \d {\d x} } {a_i} b_i + a_i \map {\frac \d {\d x} } {b_i} }$

$\ds $

$=$

$\ds \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i} b_i + \sum_{i \mathop = 1}^n a_i \map {\frac \d {\d x} } {b_i}$

$\ds $

$=$

$\ds \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$

Definition of Dot Product

$\blacksquare$