Fulton and Harris 8.30 begins:
If $\alpha$ and $\beta$ are arcs in $G$ and $\gamma(t) = \alpha(t) \cdot \beta(t)$, then
$$ \gamma(t)' = dm_{\alpha(t)}(\beta'(t)) + dn_{\beta(t)}(\alpha'(t))$$ where for any $g \in G$, $m_g: G \to G$ is left multiplication by $g$, and $n_g$ is right multiplication by $g$.
I have been very stuck on proving this product rule, and have not been able to find a good stack exchange post for it.
I have tried using the chain rule on manifolds to attack this problem, but it got a little confusing toward the middle for me. I also tried just directly evaluating both sides on an arbitrary smooth map $f: G \to \mathbb{R}$ to see if the differentials are the same as maps.
Evaluating at $t_0$, I got on the left side:
$$ \gamma'(t_0)(f) = D_t(f(\alpha(t) \cdot \beta(t))(t_0)$$
but on the right I got:
$$ \left( d(m_{\alpha(t_0)})_{\beta(t_0)}(\beta'(t_0)) + d(n_{\beta(t)})_{\alpha(t_0)}(\alpha'(t))\right)(f) = D_t(f(\alpha(t_0) \cdot \beta(t)))(t_0) + D_t(f(\alpha(t) \cdot \beta(t_0)))(t_0)$$
It is not immediately obvious to me at all that these two sides should be equal.
Am I misunderstanding the product rule I am supposed to prove?
Am I misunderstanding how to apply to right hand side to a smooth function?
If I am correct, why are the two sides equal?
Is there a better proof than simply trying to evaluate both sides on a function?
The equality of the two sides $$D_t(f(\alpha(t) \cdot \beta(t))(t_0)$$ and $$D_t(f(\alpha(t_0) \cdot \beta(t)))(t_0) + D_t(f(\alpha(t) \cdot \beta(t_0)))(t_0)$$ is immediate by the chain rule. Indeed, writing $g(s,t)=f(\alpha(s)\cdot\beta(t))$, the multivariable chain rule tells us $$D_t(g(t,t))(t_0)=D_t(g(t_0,t))(t_0)+D_t(g(t,t_0))(t_0)$$ and that is exactly the equation you want.
Alternatively, you can prove the original statement directly using the chain rule. Let $\mu:G\times G\to G$ be the multiplication map, so $\gamma(t)=\mu(\alpha(t),\beta(t))$. By the chain rule, $$\gamma'(t)=\frac{\partial\mu}{\partial x}(\alpha(t),\beta(t))\alpha'(t)+\frac{\partial\mu}{\partial y}(\alpha(t),\beta(t))\beta'(t)$$ where the partial derivatives are the derivatives of $\mu$ with respect to its first and second inputs, respectively. Now $\frac{\partial\mu}{\partial x}(x,y)$ is just $dn_y(x)$: we are differentiating the function $\mu(x,y)=n_y(x)$ with respect to $x$. Similarly, $\frac{\partial\mu}{\partial y}(x,y)$ is $dm_x(y)$. So, plugging these into the equation above, we obtain $$\gamma'(t)=dn_{\beta(t)}(\alpha(t))\alpha'(t)+dm_{\alpha(t)}(\beta(t))\beta'(t).$$ This differs from the formula you want only in that I am explicitly writing which points we are taking the differentials $dn_{\beta(t)}$ and $dm_{\alpha(t)}$, namely at $\alpha(t)$ and $\beta(t)$ respectively.