$d(f)(a) =d(g)(a)$ if $f^\prime(a)=g^\prime(a)$ for a linear operator $d$ Let $E=\mathcal C^\infty(\mathbb R,\mathbb R)$ be the space of indefinitely differentiable real functions. Suppose that $d$ is a linear operator of $E$ such that 
$$d(fg)=f d(g)+ g d(f)$$ for all $f,g \in E$.
Suppose that $f^\prime(a)=g^\prime(a)$ for a given $a \in \mathbb R$. How to prove that $d(f)(a) =d(g)(a)$?
I tried to use the similar relation of the derivative than the one of $d$ and the linearity of $d$... without success so far. Any idea?
 A: Using Hadamard's lemma, there exists $h_1,h_2\colon\mathbb{R}\rightarrow\mathbb{R}$ smooth maps such that $h_1(a)=f'(a)$, $h_2(a)=g'(a)$ and which satisfy:
$$\begin{align}f(x)&=f(a)+(x-a)h_1(x),\\g(x)&=f(a)+(x-a)h_2(x).\end{align}$$
By linearity of $d$ and since $d(1)=d(1\times 1)=d(1)+d(1)$, one has that $d$ is zero on constant maps. Therefore, one gets:
$$\begin{align}d(f)(a)&=h_1(a)d(x-a),\\d(g)(a)&=h_2(a)d(x-a).\end{align}$$
Whence the result, since $h_1(a)=f'(a)=g'(a)=h_2(a)$.

I would like to give more insight on this exercise:
Let $M$ be a smooth manifold, linear maps $d$ on $C^\infty(M,\mathbb{R})$ which satisfy:
$$d(fg)=fd(f)+gd(g)$$
are called global derivations on $M$. Using the multivariate  Hadamard's lemma, one can show that for all derivation $d$ on $M$ there exists a unique vector field $X$ such that:
$$d(f)(a)=T_af\cdot X(a).$$
This is precisely what we have shown in dimension $1$ on $\mathbb{R}$, $X(a)=d(x-a)$.
As I see that you are French, if you are interested in learning more about global derivations, I recommend reading Chapter 3 of Introduction aux variétés différentielles by J. Lafontaine.
