Let $a,b:\mathbb{R}\rightarrow M$ be two smooth curves in a manifold $M$.
We define the product of functions $\cdot$ in the usual way so that $C^\infty(\mathbb{R}\rightarrow M)$ is a monoid.
The derivative of a curve $c:\mathbb{R}\rightarrow M$ is
$$ \frac{d}{dt}\Big|_t c(t) := c_{*,t}(\frac{d}{dt}\Big|_t) $$
I would guess that
$$ \frac{d}{dt}\Big|_t (a\cdot b)(t) = \frac{d}{dt}\Big|_t a(t)\cdot b(t) + a(t) \cdot \frac{d}{dt}\Big|_t b(t) $$
I'm trying to prove this.
We have (with $m(a,b):=a\cdot b$)
$$ \frac{d}{dt}\Big|_t (a\cdot b)(t) = (a\cdot b)_{*,t}(\frac{d}{dt}\Big|_t)= \left[m_{*,(a(t),b(t))}\left( a_{*,t}, b_{*,t}\right) \right](\frac{d}{dt}\Big|_t)=m_{*,(a(t),b(t))}\left( \frac{d}{dt}\Big|_t a(t), \frac{d}{dt}\Big|_t b(t)\right)$$
First, I was thinking on using the derivative of the multiplication map like in Differential of the multiplication and inverse maps on a Lie group, but of course this is nonsensical, since $m$ is not a (smooth) group multiplication, but simply a monoid multiplication. What now, then?