Show if $U$ is an open set in $\mathbb R^p$ containing the line segment from $a$ to $b$ and if $F:U \to \mathbb R^q$ is a differentiable function on $U$, then for each vector $u \in \mathbb R^q$ there is a point $c$ on the line segment from $a$ to $b$ such that $$u \cdot (F(b)-F(a)) = u \cdot dF(c)(b-a).$$
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This is just the mean value theorem applied to the scalar valued $f(x) = u \cdot F(x)$. – copper.hat Feb 28 '20 at 04:03
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While looking for duplicate questions, I found the answer as an aside on a different question here.
Choose $u \in \mathbb R^q$. Let $$G(t):=u\cdot F(a+t(b-a))$$ on the domain $[0,1]$. Note that $\{a+t(b-a):t\in [0,1]\}$ is the line segment from $a$ to $b$, which is contained in $U$. Also notice $$dG(t) = u\cdot dF(a+t(b-a))(b-a).$$ From the Mean Value Theorem for real-valued functions, there is a $t_0\in (0,1)$ such that $$G(1)-G(0)=dG(t_0)(1-0).$$
Notice $G(1)=u\cdot F(b)$ and $G(0)=u\cdot F(a)$. Let $c=a+t_0(b-a)$. Then we have $$u\cdot F(b)-u\cdot F(a) = u\cdot df(c)(b-a)$$ which can be written as $$u\cdot (F(b)-F(a)) = u\cdot df(c)(b-a).$$
Jacob
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