Derivative of a group action Let $\phi : G \times M \rightarrow M$ be a group action on a smooth manifold $M$ and Lie group $G$.
Then we define $$f(t):=\phi(g(t),d(t)).$$
where $g: I \rightarrow G$ and $d: I \rightarrow M$ are smooth maps.
Now I'd say: $$f'(t) = D\phi(g(t),d(t))(g'(t),d'(t)).$$
In the lecture we wrote: $$f'(t) = D \phi_{g(t)} d(t) d'(t) + D \phi_{g(t)} (d(t))  X_{\zeta(t)} (d(t))$$ where $\zeta(t) = dL_{g(t)^{-1}}g'(t)$ is an element of the Lie Algebra, $dL$ is the derivative of the left-translation and $X_{\zeta}(p):=\frac{d}{dt}|_{t=0} \phi_{e^{t \zeta}}(p)$ for any $\zeta$ in the Lie Algebra.
The thing is that I don't have the slightest clue, how one could come up with this second expression. Could anybody explain to me whether I am wrong or give me an idea why both expressions might be equivalent?
 A: So if you had no idea how it worked, why didn't you ask in the lecture?
First, split the problem up (I will use the letter $u$ instead of $d$ because of possible confusion with the differential) into the question how to differentiate the function $f(t) = \phi(g, u(t)) = g \cdot u(t)$ for fixed $g \in G$, and how to differentiate the function $f(t) = \phi(g(t), p) = g(t) \cdot p)$ for a fixed point $p \in M$.
The first is really straightforward. $g$ acts as a diffeomorphism $\phi_g$ of $M$, so we just have to calculate
$$ f^\prime(t) = \frac{\partial}{\partial t} \bigl\{ \phi_g(u(t))\bigr\} = D \phi_g\big|_{u(t)} u^\prime(t)$$
Here $D\phi_g$ is just the differential of the diffeomorphism $\phi_g$, and it has to be evaluated at the point $u(t)$.
Now the second case can be reduced to the first case. Namely, suppose you want to differentiate at the point $t_0$ and write $h_t := g_{t_0}^{-1} g_t$ and $u(t) = g_{t_0}^{-1} g_t \cdot p$. Then clearly $g_t \cdot p = g_{t_0} \cdot u(t)$, so that 
$$f^\prime(t_0) = D\phi_{g_{t_0}}\big|_{u(t_0)} u^\prime(t_0) = D\phi_{g_{t_0}}\big|_{p} u^\prime(t_0).$$
It remains to calculate $u^\prime(t)$. We have
$$h^\prime(t_0) = \frac{\partial}{\partial t}\Big|_{t=t_0} g_{t_0}^{-1} g_t = dL_{g_{t_0}^{-1}} g^\prime(t) = \xi(t_0)$$
where $\xi(t_0)$ is an element of the Lie algebra $\mathfrak{g} = T_eG$. Hence $h_t = g_{t_0}^{-1}g_t$ is a curve starting at $e \in G$ at time $t_0$ with velocity $\xi$. $\exp(t\xi)$ is another such curve. Hence
$$u^\prime(t_0) = \frac{\partial}{\partial t}\Big|_{t=t_0}g_{t_0}^{-1}g_t\cdot p = \frac{\partial}{\partial t}\Big|_{t=t_0} e^{t\xi}\cdot p = X_\xi(p)$$
The claim follows.
