= Solution
The scalar-valued multivariable <mean value theorem> states that, if $f$ is differentiable on an <open set> containing the closed line segment from $x$ to $y$, then for $x\ne y$ there exists $s\in(0,1)$ such that
$$
\boxed{f(y)-f(x)=Df(x+s(y-x))(y-x)=\nabla f(x+s(y-x))\cdot(y-x).}
$$
This follows by applying the one-dimensional <mean value theorem> to $t\mapsto f(x+t(y-x))$ on $[0,1]$. The segment condition is necessary for this formulation on a possibly nonconvex domain.
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