These representation theory notes leave the following claim to the reader:
Recall that a derivation on an algebra is a map $d$ such that $d(ab)=d(a)b+ad(b)$. If $A$ is a finite-dimensional algebra over $\Bbb{R}$ or $\Bbb{C}$, $\gamma:\Bbb{R}\to \operatorname{Aut}_\Bbb{R}(A)$ is differentiable in some neighborhood of $0$, and $\gamma(0)=\operatorname{id}_A$, then $\gamma'(0):A\to A$ is a derivation.
This sounds a lot like the two pictures of the tangent space to a manifold at some point $p$ we get by considering smooth paths through $p$ and derivations of the algebra of germs of smooth real-valued functions at $p$, but that correspondence doesn't appear to me to actually go through.
Specifically, I think that the path of automorphisms of $\Bbb{R}$ given by multiplication by $e^t$ is a counterexample: it's smooth and the identity at $0$, but its derivative at $0$ is $1$, while $1(ab)\neq 1(a)b+a1(b)=2ab$ when $ab\neq 0$. Am I missing something?