Computing the differential of a Lie group action 
(Ex. 27.4 page 252 Loring Tu)     (The differential of an action). Let $\mu: P \times G \rightarrow P$. For $g \in G$, the tangent space $T_gG$ may be identified with $l_{g*} \mathfrak{g} $, where $l_g:G \rightarrow G$ is left multiplication by $g \in G$ and $\mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P \times G$ is of the form 
      $$(X_P, l_{g*} A)$$
  for $X_p \in T_pP$ and $A \in \mathfrak g$. The differential is given by 
      $$ \mu_*= \mu_{*,(p,g)} :T_{(p,g)}(P \times G) \rightarrow T_{pg} P$$ 
  is given by 
      $$ \mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + \underline{A}_{pg} $$ 



Definition: $\underline{A}$ is the fundamental vector field on $P$ associated to $A \in \mathfrak{g}$, 
  $$ \underline{A}_p  = \frac{d}{dt}\Big|_{t=0} p \cdot e^{tA} \in T_pP$$


I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out. 
 A: So with my edits above I think I have came up with the answer: 
$$ \cdots \cdots \cdots $$ 
Let us identify 
$$T_{(p,g)}(P \times G) \cong T_pP \oplus T_gG$$ 
 hence by linearity of differential, it suffices to compute each component. 
$$ \cdots \cdots \cdots $$ 
Let $(X_p, 0) \in T_pP \oplus T_gG$. $\varphi(t):(-\varepsilon, \varepsilon) \rightarrow  P$  be a smooth curve, $\varphi(0) = p, \varphi'(0)=X_p$. Define $$\gamma(t):(-\varepsilon, \varepsilon) \rightarrow P \times G, \quad t \mapsto (\varphi(t), g)$$
Then
\begin{align*}
\mu_*(X_p,0) &= \mu_* \gamma_* (\frac{d}{dt} \Big|_{t=0}) \\ 
& =(r_g \circ \varphi)_* (\frac{d}{dt} \Big|_{t=0}) \\
&= (r_g)_* X_p 
\end{align*}
$$ \cdots \cdots \cdots $$ 
To compute $(0,(l_g)_*A)$ we consider the curve, 
$$ \gamma(t): (-\varepsilon, \varepsilon) \rightarrow P \times G, (p, g \cdot e^{tA}) $$ 
Then we have 
\begin{align*}
\mu_* \gamma_* (\frac{d}{dt} \Big|_{t=0}) &= (c_{pg})_*(\frac{d}{dt} \Big|_{t=0}) \\ 
&=: \underline{A}_{pg} \end{align*}
where $c_{pg}(t):\Bbb R \rightarrow P$ is the map given by $pg \cdot e^{tA}$.
