5c287c…: “As regards a quiet cube”

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Suppose $\vec {p} : \R \to \R ^3$ and $\vec {q} : \R \to \R ^3$ are vector-valued functions.

Further assume $\vec {p}'(-1) = \vector {1,2,3}$ and $\vec {q}'(-1) = \vector {1,1,-2}$ .

Let $\vec {v}(t) = \vec {p}(t) + \vec {q}(t)$ be yet another vector-valued function.

Then $\vec {v}'(-1) = \vector {\answer {2},\answer {3},\answer {1}}$ .