Chain rule for functions of d variables I'm reading up this proof and get stuck at the step where the author derives the formula for $g'(t)$. Could anyone write out the steps involved?


 A: Let $z:=x+t(y-x)$, so that $z_i=x_i+t(y_i-x_i)$.
We have
$$g(t)=f(z),$$
$$g'(t)=\frac{dg(t)}{dt}=
\frac{\displaystyle\sum_i\dfrac{\partial f(z)}{\partial z_i}{dz_i}}{dt}=\sum_i\frac{\partial f(z)}{\partial z_i}\frac{dz_i}{dt}=\sum_i\frac{\partial f(z)}{\partial z_i}(y_i-x_i).$$
A: Sure. $\newcommand{\R}{\mathbb{R}}$
You have 
$$
g'(t) = \frac{d}{dt} g(t) = \frac{d}{dt} f(x + t(y-x)),
$$
where we need the chain rule for the last expression since it is a composition of two functions. Let me denote with $D$ the derivative (differential) then in general the chain rule for $\varphi : \R^n \to \R^m$ and $\psi : \R^l \to \R^m$, where $\varphi$ and $\psi$ are $C^1$ is 
$$
D \phi \circ \psi(x) = (D\varphi)(\psi(x))\circ (D\psi)(x).
$$
Notice that for a 1-d function $t \mapsto g(t)$ $D g = \frac{d}{dt}g$.
Now in your case we get:
$$
g'(t) = \frac{d}{dt} f(x + t(x-y) = (Df)(x+t(y-x))\circ \frac{d}{dt} (x+t(y-x)).
$$
I assume you have heard of the jacobian. Then the linear map $Df$ can be associated with that jacobian matrix and the composition $\circ$ acutally becomes matrix multiplication. So
$$
Df = (\frac{\partial}{\partial x_1} f, \dots, \frac{\partial}{\partial x_d} f) =: (f_{x_1} \dots f_{x_d}),
$$
which is a $1\times d$ matrix and 
$$
\frac{d}{dt} (x+t(y-x)) = (y -x)
$$
which is a $d\times 1$ matrix. So doing this matrix product gives you
$$
(Df)(x+t(y-x))\circ \frac{d}{dt} (x+t(y-x)) = \big(f_{x_1} \dots f_{x_d}\big)(x+t(y-x)) \cdot (y-x) 
= \sum_{i=1}^d f_{x_i}(x+t(y-x)) \cdot(y_i - x_i),
$$
where this last $\cdot $ is actually just the usual multiplication in the real numbers.
