Revisiting the product rule for derivatives 
Let $E=C^{\infty}(\mathbb R, \mathbb R)$
Consider a linear transformation on $E$: $\delta$ such that $\forall f, g \in E, \delta(fg) =g\delta(f) +f\delta(g)$
Prove that there is some $u\in E$ such that $\forall f\in E, \delta(f) =uf'$

My answer is that $u:x\to \delta(Id)(0)$ where $Id$ denotes the identity function.
It is very easy to prove that $\delta$ annihilates constant functions.
Consider a fixed $x_0$ and  these few functions :
$\displaystyle g:x \to \frac{f(x_0+x)-f(x_0)}{x}$
$ \displaystyle h:x \to f(x_0+x)-f(x_0)$
$ \displaystyle m:x \to f(x_0+x)$
All these functions belong to $E$.
Let $x\in \mathbb R$
$\delta(Id\times g) (x) =x\delta(g(x)) + g(x) \delta(Id) (x) $
Hence
$\delta(h) (x) =\delta(m) (x) =x\delta(g(x)) + g(x) \delta(Id) (x) $
Now, let $x\to 0$
The RHS becomes $f'(x_0) \delta(Id) (0)$ since $\delta(g) $ is bounded around $0$.
Question
But why is $\displaystyle \lim_{x\to 0}\delta(m)(x)=\delta(f)(x_0)$?
It is very intuitive, but I fail finding a proof.
 A: One can verify that for any $u \in E$, the rule $\delta(f) = uf'$ defines a derivation of $E$, i.e. an $\mathbb{R}$-linear map $\delta: E \to E$ with the property that $\delta(fg) = g\delta(f) + f\delta(g)$. It is not necessary that the coefficient $u$ be a real constant, as the definition "$u : x \mapsto \delta(\operatorname{Id})(0)$" in your post suggests. The function $u$ will be constant if and only if $\delta$ commutes with all operators $T_a : E \to E$, where for $a \in \mathbb{R}$, we define $T_a(f)(x) = f(x+a)$. In fact, the last formula in your post claims that $\delta(T_{x_0}f)(0) = T_{x_0}(\delta f)(0)$, which is not true in general.
Assuming that $\delta(f) = uf'$ for some $u \in E$ and putting $f = \operatorname{Id}$ in this equality, we obtain $u = \delta(\operatorname{Id})$ (no evaluation at $x=0$). One possible way to prove that this indeed is the case, is as follows. Fix $x_0 \in \mathbb{R}$ and consider the Taylor formula with integral form of the remainder
$$ f(x) = f(x_0) + f'(x_0)(x-x_0) + R(x)(x-x_0)^2, \quad \text{where} \quad R(x) = \int_{0}^{1} (1-t)\,f''((1-t)x_0 + tx)\,dt.$$
Since $f \in E$, it is possible to differentiate the integrand in the definition of $R$ any number of times, to show that $R \in E$, as well. Now, apply $\delta$ to both sides of Taylor's formula, and then put $x=x_0$ in the resulting equality. This would give us the relation $\delta(f)(x_0) = f'(x_0)\delta(\operatorname{Id})(x_0)$. Since this is valid for any $x_0 \in \mathbb{R}$, we conclude that $\delta(f) = \delta(\operatorname{Id})\cdot f'$.
