proof of chain rule (several variables)

We first consider the case $m=1$ i.e. $G:I\to {ℝ}^n$ where $I\subset ℝ$ is a neighbourhood of a point $x_0\in ℝ$ and $F:U\subset {ℝ}^n\to ℝ$ is defined on a neighbourhood $U$ of $y_0=G(x_0)$ such that $G(I)\subset U$. We suppose that both $G$ is differentiable at the point $x_0$ and $F$ is differentiable in $y_0$. We want to compute the derivative of the compoundfunction $H(x)=F(G(x))$ at $x=x_0$.

By the definition of derivative (using Landau notation) we have

$$F(y_0+k)=F(y_0)+DF(y_0)k+o(|k|).$$

Choose any $h\ne 0$ such that $x_0+h\in I$ and set $k=G(x_0+h)-G(x_0)$ to obtain

${\displaystyle \frac{H(x_0+h)-H(x_0)}{h}}$$={\displaystyle \frac{F(G(x_0+h))-F(G(x_0))}{h}}$

$={\displaystyle \frac{F(G(x_0)+k)-F(G(x_0))}{h}}={\displaystyle \frac{F(y_0+k)-F(y_0)}{h}}$

$={\displaystyle \frac{DF(y_0)(G(x_0+h)-G(x_0))+o(|G(x_0+h)-G(x_0)|)}{h}}$

$=DF(y_0){\displaystyle \frac{G(x_0+h)-G(x_0)}{h}}+{\displaystyle \frac{o(|G(x_0+h)-G(x_0)|)}{h}}.$

Letting $h\to 0$ the first term of the sum converges to $DF(y_0)G^{\prime }(x_0)$ hencewe want to prove that the second term converges to $0$.Indeed we have

$$\left|\frac{o(|G(x_0+h)-G(x_0)|)}{h}\right|=\left|\frac{o(|G(x_0+h)-G(x_0)|)}{|G(x_0+h)-G(x_0)|}\right|\cdot \left|\frac{G(x_0+h)-G(x_0)}{h}\right|.$$

By the definition of $o(\cdot )$ the first fraction tends to $0$, whilethe second fraction tends to the absolute value of $G^{\prime }(x_0)$. Thus the producttends to $0$, as needed.

Consider now the general case $G:V\subset {ℝ}^m\to U\subset {ℝ}^n$.Given $v\in {ℝ}^m$ we are going to compute the directional derivative

$$\frac{\partial F\circ G}{\partial v}(x_0)=\frac{dF\circ g}{dt}(0)$$

where $g(t)=G(x_0+tv)$ is a function of a single variable $t\in ℝ$. Thuswe fall back to the previous case and we find that

$$\frac{\partial F\circ G}{\partial v}(x_0)=DF(G(x_0))g^{\prime }(0).=DF(G(x_0))\frac{\partial G}{\partial v}(x_0)$$

In particular when $v=e_k$ is the $k$-th coordinate vector, we find

$$g^{\prime }(0)=D_{x_k}F\circ G(x_0)=DF(G(x_0))D_{x_k}G(x_0)=\sum _{i=1}^n{D}_{y_i}G(x_0)D_{x_k}{G}^i(x_0)$$

which can be compactly written

$$DF\circ G(x_0)=DF(G(x_0))DG(x_0).$$