Derivative of Sum of Vector-Valued Functions

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Theorem

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


Then the derivative of $\map {\mathbf v} t = \map {\mathbf a} t + \map {\mathbf b} t$ is given by:

$\dfrac {\d \mathbf v} {\d t} = \map {\dfrac \d {\d t} } {\mathbf a + \mathbf b} = \dfrac {\d \mathbf a} {\d t} + \dfrac {\d \mathbf b} {\d t}$


Proof

\(\ds \dfrac {\d \mathbf v} {\d t}\) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf v} {t + h} - \map {\mathbf v} t} h\) Definition of Derivative of Vector-Valued Function
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\paren {\map {\mathbf a} {t + h} + \map {\mathbf b} {t + h} } - \paren {\map {\mathbf a} t + \map {\mathbf b} t} } h\) Definition of $\mathbf v$
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\paren {\map {\mathbf a} {t + h} - \map {\mathbf a} t} + \paren {\map {\mathbf b} {t + h} - \map {\mathbf b} t} } h\) rearranging
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {t + h} - \map {\mathbf a} t} h + \lim_{h \mathop \to 0} \dfrac {\map {\mathbf b} {t + h} - \map {\mathbf b} t} h\)
\(\ds \) \(=\) \(\ds \dfrac {\d \mathbf a} {\d t} + \dfrac {\d \mathbf b} {\d t}\) Definition of Derivative of Vector-Valued Function

$\blacksquare$


Sources

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