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
- 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: $(3.3 a)$