Show that any linear map $T: C^{\infty}(\mathbb{R}) \to \mathbb{R}$ satisfying
- (a) $T(1)=0$,
- (b) $T(x)=1$,
- (c) $T(fg) = f(0)T(g)+g(0)T(f)$
is unique, and thus show that any such $T$ satisfies $T(f)=f'(0)$.
Now, I have proved that $T $ must be unique but I am not able to prove that $T(f)$ must be equal to $f'(0)$. I tried by putting $g=1 $ in $T(fg) = f(0)T(g)+g(0)T(f)$ and and then differentiating , differentiating $T(fg) = f(0)T(g)+g(0)T(f)$ and then putting $g=1$ but they don't help with what I want to prove.
So, what approach should I use.