Derivative of Square of Vector-Valued Function

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Theorem

Let $\mathbf a: \R \to \R^n$ be a differentiable vector-valued function.


The derivative of its square is given by:

$\map {\dfrac \d {\d x} } {\mathbf a^2} = 2 \mathbf a \cdot \dfrac {\d \mathbf a} {\d x} = 2 a \dfrac {\d a} {\d x}$

where $a = \norm {\mathbf a}$ is the magnitude of $\mathbf a$.


Proof

\(\ds \map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b}\) \(=\) \(\ds \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}\) Derivative of Dot Product of Vector-Valued Functions
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf a}\) \(=\) \(\ds \dfrac {\d \mathbf a} {\d x} \cdot \mathbf a + \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}\) setting $\mathbf a = \mathbf b$
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d x} } {\mathbf a^2}\) \(=\) \(\ds \dfrac {\d \mathbf a} {\d x} \cdot \mathbf a + \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}\) Definition of Square of Vector Quantity
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d x} } {\mathbf a^2}\) \(=\) \(\ds 2 \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}\) Dot Product Operator is Commutative
\(\ds \) \(=\) \(\ds 2 a \dfrac {\d a} {\d x}\) Dot Product of Vector-Valued Function with its Derivative

$\blacksquare$


Sources

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