If $\phi_1$ and $\phi_2$ are two paths in a Lie group, what is the derivative $(\phi_1 \times \phi_2)^\prime(0)$? Let $\phi_1, \phi_2$ be two paths in a Lie group $G$ with $\phi_i(0) = g_i \in G$ and $\phi^\prime(0) = X_i \in \frak{g}$. Denote $\psi(t) = \phi_1(t) \phi_2(t)$. We get $\psi(0) = g_1 g_2$. What is $\psi^\prime(0)$?
By definition, it is the differential defined for any smooth $f : G \to \mathbb{R}$ by
\begin{eqnarray*}
\psi^\prime(0)(f) &=& \frac{d}{dt}\left(f \circ \psi(t)\right)|_{t = 0} \\
&=& \frac{d}{dt}\left(f \circ (\phi_1(t) \phi_2(t))\right)|_{t = 0} \\
&=& \frac{d}{dt} \left(f ( L_{\phi_1(t)} \phi_2(t) \right)|_{t = 0} \\
\end{eqnarray*}
I do not see how to proceed from there and see how the derivation looks like.
 A: Let $m : G\times G \to G$ be the multiplication map. Then $\psi (t) = m(\phi_1(t), \phi_2(t))$. By chain rule, 
$$ \psi'(0) = m_* (\phi_1'(0), \phi'_2(0)),$$
where $m_* : T_{g_1}G \times T_{g_2} G \to T_{g_1g_2} G$ is the derivative of $m$ at $(g_1, g_2)$. The differential of $m$ can be calculated easily (see here). In particular since $m_*$ is bilinear, we have 
$$m_* (\phi_1(0), 0) = \frac{d}{dt} \phi_1(t)g_2 \bigg|_{t=0} = (r_{g_2})_* X_1,$$
$$m_* (0, \phi_2(0)) = \frac{d}{dt} g_1\phi_2(t)\bigg|_{t=0} = (\ell_{g_2})_* X_2.$$
Thus
$$\psi'(0) = (r_{g_2})_*X_1 + (\ell_{g_1})_* X_2.$$
One can write down everything as if they are in the Lie algebra $\mathfrak g$: Let say we are identifying each $T_gG$ with $\mathfrak g : = T_eG$ by left multiplication 
$$\tag{1} (\ell_g)_* : \mathfrak g \to T_gG.$$
Then $X_1, X_2$ are really $(\ell_{g_1})_* Y_1$ and $(\ell_{g_2})_* Y_2$ for some $Y_1, Y_2 \in \mathfrak g$ respectively. Then 
$$\begin{split} 
\psi'(0) &= (r_{g_2})_*X_1 + (\ell_{g_1})_* X_2 \\
&= (r_{g_2})_*(\ell_{g_1})_* Y_1 + (\ell_{g_1})_* (\ell_{g_2})_* Y_2  \\
&= (\ell_{g_1})_* (r_{g_2})_* Y_1 +(\ell_{g_1g_2})_* Y_2 \\
&= (\ell_{g_1})_* (\ell_{g_2})_* (\ell_{g_2^{-1}})_* (r_{g_2})_* Y_1 +(\ell_{g_1g_2})_* Y_2 \\
&= (\ell_{g_1g_2})_* (\text{Ad}_{g_2^{-1}} Y_1 + Y_2)
\end{split}$$
Thus we have 
$$\psi'(0) = \text{Ad}_{g_2^{-1}} Y_1 + Y_2$$ 
under the identification $(1)$. Here $\text{Ad}_g : \mathfrak g \to \mathfrak g$ is the Adjoint representation, which is by definition the differential of the conjugate mapping 
$$ G \to G , \ \ \ h \mapsto g hg^{-1}.$$
