Conclude the "MVT for scalar fields"- proof

Question

So, e.g $f=a^2+b^2$ (red function in the picture). For two points $x,y$ $\in$ $(-3,3)\times(-3,3)$ we define $g$ as $g: f(ty+(1-t)x)$, and that will be the gray curve. MVT says, there is a $t \in(0,1)$ such that $g(1)-g(0)=g'(t)$, so $f(y)-f(x) = g'(t)$. I want to show the MVT for scalar fields, means there is a $\xi$ with $f(y)-f(x) = \langle \mathrm{grad}f(\xi), y-x\rangle.$ How to transform $g'(t)$ $\textbf{formally}$ correct into the scalar product I need, like $g'(t) = \ldots = \langle \cdot , \cdot \rangle$?

Answer 1

If $\mathbf x=(x_1,x_2),\mathbf y=(y_1,y_2)$ then $t\mathbf y+(1-t)\mathbf x=\left(x_1+(y_1-x_1)t,x_2+(y_2-x_2)t\right)$.

So $$g(t)=(x_1+(y_1-x_1)t)^2 + (x_2+(y_2-x_2)t)^2$$

So $$\begin{align}g'(t)&=2(x_1+(y_1-x_1)t)(y_1-x_1)+2(x_2+(y_2-x_2)t)(y_2-x_2)\\ &=2(x_1(y_1-x_1)+x_2(y_2-x_2))+2t\left((y_1-x_1)^2+(y_2-x_2)^2\right)\\ &=2\langle \mathbf x,\mathbf y-\mathbf x\rangle +2t\langle \mathbf y-\mathbf x,\mathbf y-\mathbf x\rangle\\ &=2\langle t\mathbf y+(1-t)\mathbf x,\mathbf y-\mathbf x\rangle \end{align} $$