About linear compositions on definition of differentiability The differentiability of $f:\mathbb{R}^n\to\mathbb{R}^m$ at a point $x$ is that there exists a linear transformation $T:\mathbb{R}^n\to\mathbb{R}^m$ such that $lim_{h\to 0}\frac{1}{\|h\|}\|f(x+h)-f(x)-T(h)\|=0$.
Also, the chain rule is given by $$d(G\circ F)(a) = dG(F(a))\circ dF(a),$$
by the usual assumptions. Now, I want to prove the quotient rule. For this, I need to show that $$d(1/G)=\frac{-dG(a)}{G^2(a)}.$$
Indeed, I defined $\phi(x)=\frac{1}{x}$ a real function and $G:\mathbb{R}^n\to\mathbb{R}$. Then, by chain rule,
$$d(\phi\circ G)(a) = d\phi(G(a))\circ dG(a)=\frac{-1}{G^2(a)}\circ dG(a).$$
But how do I get the desired result? Moreover, how does this composition works?: $$\frac{-1}{G^2(a)}\circ dG(a).$$
$dG(a)$ is a linear map from $\mathbb{R}^n\to\mathbb{R}$, and $\frac{-1}{G^2(a)}$ goes from $\mathbb{R}^n$ to $\mathbb{R}$ as well. So how do I get the product? I'm missing something that I don't see it.
 A: Given a function $F : A \subseteq \Bbb{R}^n \to \Bbb{R}^m$, and a point $a$ in the interior of the domain $A$, the derivative $(dF)(a)$, if it exists, is a linear map between $\Bbb{R}^n$ and $\Bbb{R}^m$. It would make sense, therefore, to consider $((dF)(a))(x)$, the result of applying this linear map to some $x \in \Bbb{R}^n$.
So, if we consider the function $F : \Bbb{R} \setminus \{0\} \to \Bbb{R}$ in this same way, we need to think about $(dF)(a)$ less as a scalar (which would be $-1/a^2$), and more as a linear map from $\Bbb{R}$ to $\Bbb{R}$ (which would be $x \mapsto -x/a^2$). I think this is what's tripping you up.
If we have $G : \Bbb{R}^n \to \Bbb{R}$ is differentiable and non-zero at $a$, then chain rule says that
$$d(1/G) = d(F \circ G) = ((dF) \circ G)(a) \circ ((dG)(a)).$$
The $((dF) \circ G)(a)$ term could also be written as $(dF)(G(a))$. This is a linear map, taking $x$ and mapping it to $x / (G(a))^2$. So, the total map is:
$$x \in \Bbb{R}^m \overset{(dG)(a)}\mapsto ((dG)(a))(x) \in \Bbb{R} \overset{(dF)(G(a))}\mapsto -\frac{((dG)(a))(x)}{(G(a))^2} \in \Bbb{R}.$$
In other words, $d(1/G)$ is the linear map $(dG)(a) : \Bbb{R}^n \to \Bbb{R}$, but divided by the scalar $-G(a)^2$.
Hope that helps.
