Frechet derivative chain rule I need to prove that, 

Fix $U\subset \mathbb{R}^m, V\subset \mathbb{R}^n$ open, $f =
 (f_1,\cdots,f_n):U\to\mathbb{R}^n$ such that $f(U)\subset V$ and each
  coordinate function $f_k:U\to\mathbb{R}$ is differentiable at the
  point $a\in U$. Let $g:V\to \mathbb{R}$ be a differentiable function
  at the point $b=f(a)$. Then, the composite $g\circ f:U\to\mathbb{R}$
  is differentiable in $a$ and its partial derivatives are given by:
$$\frac{\partial(g\circ f)}{\partial x_i}(a) =
 \sum_{k=1}^n\frac{\partial g}{\partial y_k}(b)\cdot \frac{\partial
 f_k}{\partial x_i}(a)$$

Here's the demonstration in my book:
Let $U_0$ be the set of vectors $v = (\alpha_1,\cdots,\alpha_n)\in\mathbb{R}^m$ such that $a+v\in U$. For $v\in U_0$ and $k=1,\cdots,n$ we have:
$$f_k(a+v) = f_k(a)+\sum_{i=1}\frac{\partial f_k}{\partial x_i}\cdot \alpha_i + p_k\cdot |v|$$
where each $p_k = p_k(v)$ is a function defined in $U_0$, continuous in point $0$, which annihilates when $v=0$
(We'll consider $\frac{\partial f_k}{\partial x_i}$ and $\frac{\partial g}{\partial y_k}$ to be in the points $a$ and $b$ respectively).
Consider the application $w = (\beta_1,\cdots,\beta_n):U_0\to\mathbb{R}^n$, continuous in $0$ such that the coordinate functions are defined by:
$$\beta_k(v) = \sum_{i=1}^m\frac{\partial f_k}{\partial x_i}\cdot \alpha_i + p_k\cdot |v|$$
Adopting, for example, the norm of the sum, we have $|\alpha_i|/|v|\le 1$ if $v\neq 0$, therefore each $|\beta_k|/|v|$, and therefore the function $|w|/|v|$, is bounded in a neighborhood of $v=0$. Writing $gf$ instead of $g\circ f$, we can say that by the formula for $f_k(a+v)$ given before, that, from differentiability of $g$ in $b=f(a)$ and that for al v\in U_0$, we have:
$$gf(a+v) = g(b+w) = g(b) + \sum_{i=1}^{n}\frac{\partial g}{\partial y_k}\cdot \beta_k + \sigma\cdot |w|$$
where $\sigma = \sigma(v)$ is a continuous real function in $0$, which anihilates in $v=0$.
Then it just concludes that this is the proof
What is that 'writing $gf$ instead of $g\circ f$' thing?
 A: Okay, I shall give a demonstration of the chain rule in a bit more general context.
Definition. A function $f,$ defined on an open subset of a normed vector space $\mathrm{V}$ and with values in another normed space $\mathrm{W}$, is differentiable at the point $a$ (in its domain) if there exists a continuous linear function $\mathrm{T}:\mathrm{V} \to \mathrm{W}$ such that $f(a + h) = f(a) + \mathrm{T}h + o(\|h\|)$ as $h \to 0,$ where, as it is standard, "$u = o(v(x))$ as $x \to x_0$" (here $u$ is a $\mathrm{W}$-valued function and $v$ is a nonnegative function) means $\displaystyle \lim_{\substack{x \to 0 \\ v(x) \neq 0}}\dfrac{\|u(x)\|}{v(x)} = 0.$
Corollary 1. If $f$ is differentiable at $a,$ its derivate is unique. (If $\mathrm{T}$ and $\mathrm{L}$ satisfy the defining property of being differentiable, one gets $(\mathrm{T} - \mathrm{L})h = o(h)$ which implies at once $\|\mathrm{T} - \mathrm{L}\| = 0$ or $\mathrm{T} = \mathrm{L}$). Whence, one denotes by $f'(a)$ the derivative of $f$ at $a$ (so, $f'(a):h \mapsto f'(a) \cdot h$ is a continuous linear function).
Corollary 2. The derivative of the restriction of linear continuous function in any interior point of its domain, is the linear function.
Remarks. 
(1) The derivative need not scalars in $\Bbb R$ or $\Bbb C,$ so most proofs
    regarding only differentiation are valid simultaneosly in the
    complexes or in the reals.
(2) Since the derivative is unique, its
    definition basically says that a function being differentiable at
    one point $a$ is the same as saying that "in the small
    neighbourhoods of $a,$ the increments of $f$ behave as a linear
    component plus an error that is infinitesimal in comparison with the
    linear part." Such approximation is unique.
(3) Therefore, one can
    bet that the best unique linear approximation of a composition is
    the composition of the best uniques linear approximations of the components; this is the "chain rule", which, by the way, loses all its trace of meaning when one forces (as your book does) an arbitrary expansion of some basis.
Chain rule. Assume $\mathrm{U}, \mathrm{V}$ and $\mathrm{W}$ are three normed spaces; let $\mathrm{E}$ be an open subset of $\mathrm{U}$ and $\mathrm{F}$ another such for $\mathrm{V}.$ Suppose $f:\mathrm{E} \to \mathrm{V}$ and $g:\mathrm{F} \to \mathrm{W}$ are differentiable at $a \in \mathrm{E}$ and $b = f(a) \in \mathrm{F},$ respectively. Then, $g \circ f,$ which is defined on some neighbourhood of $a,$ is differentiable at $a$ and its derivative is $(g \circ f)'(a) = g'(f(a)) \circ f'(a).$
Proof. In some small neighbourhood $\mathrm{B}_1$ of $a,$ the following expansion follows from definition $$f(a + h) = f(a) + f'(a) \cdot h + o_f(\|h\|) = b + f'(a) \cdot h + o_f(\|h\|).$$ Similarly, in some small neighbourhood $\mathrm{C}_1$ of $b,$ $$g(b + k) = g(b) + g'(b) \cdot k + o_g(\|k\|).$$ Since $\|f'(a) \cdot h + o_f(\|h\|)\| \leq \left(\|f'(a)\| + \left\|\dfrac{o_f(\|h\|)}{\|h\|}\right\|\right) \|h\|,$ one can restrict $h \in \mathrm{B}_2 \subset \mathrm{B}_1$ such that $k = f'(a) \cdot h + o_f(\|h\|)$ has norm $\leq (\|f'(a)\| + 1) \|h\|.$ If necessary, restrict further $h \in \mathrm{B}_3 \subset \mathrm{B}_2$ such that $b + k \in \mathrm{C}_1,$ where $k$ is as before. Therefore, for $h \in \mathrm{B}_3,$
$$\begin{array}
((g \circ f)(a + h) &= g(b + k) = g(b) + g'(b) \cdot k + o_g(\|k\|) \\
&= g(b) + g'(b) \cdot (f'(a) \cdot h + o_f(\|h\|)) + o_g(\|k\|) \\
&= g(b) + g'(f(a)) f'(a) \cdot h + g'(b) \cdot o_f(\|h\|) + o_g(\|k\|) \\
&= g(b) + g'(f(a)) f'(a) \cdot h + \psi(h),
\end{array}$$
where $\psi(h) = g'(b) \cdot o_f(\|h\|) + o_g(\|k\|)$ is defined for $h \in \mathrm{B}_3$ and needs to be shown that, when divided by $\|h\|,$ the quotient tends to zero. This is easy, because $\|k\| \leq (\|f'(a)\| + 1)\|h\|,$ so $\psi(h)$ is the sum of two $o(\|h\|)$ functions, making this of $\psi$ a $o(\|h\|)$ function too. The proof is terminated. CQFD
Corollary. If $f$ is defined in some open subset of $\Bbb R^p$ with values in $\Bbb R^q$, then, the $q \times p$ matrix that represents $f'(x)$ in the canonical basis is $\left( \dfrac{\partial f_j}{\partial x_i} \right)_{\substack{i = 1, \ldots, p\\j = 1, \ldots, q}}.$ (Since the limit is coordinatewise, it suffices to consider the case $q = 1;$ for this case consider the canonical injection $\eta_j:\Bbb R \to \Bbb R^p$ and the canonical projection $\pi_i:\Bbb R^p \to \Bbb R,$ then notice that that (1) these functions are linear and continuous, hence differentiable everywhere with them being their own derivate and (2) the identity of $\Bbb R^d$ can be written as $id = \sum\limits_{k = 1}^p \eta_k \circ \pi_k$ and apply the chain rule to $f = id \circ f$).
