Let $G$ be a Lie group and $\mathfrak{g}$ its associated Lie algebra. Define $$Ad: G \rightarrow GL(\mathfrak{g})\\ g \mapsto (c_g)_*$$ to be the adjoint representation where $$c_g: G \rightarrow G\\ x \mapsto gxg^{-1}.$$
To prove that this map is smooth, according to the book Mathematical Gauge Theory by Hamilton,
It suffices to show that for every $v \in \mathfrak{g}$ the map $$Ad(\cdot)v: G \rightarrow \mathfrak{g}$$ is smooth, because if we choose a basis for the vector space $\mathfrak{g}$, it follows that $Ad$ is a smooth matrix representation. The map $Ad(\cdot)v$ is equal to the composition of smooth maps $$G \rightarrow TG \times TG \rightarrow T(G \times G) \rightarrow TG$$ given by $$g \mapsto \big((g,0), (e,v)\big) \rightarrow \big((g,e), (0,v)\big) \mapsto D_{(g,e)}c(0,v)$$ where $$c: G\times G \rightarrow G\\ (g,x) \mapsto gxg^{-1}.$$ This implies the claim.
I am having trouble understanding the logic behind the map $Ad(\cdot)v$. How does proving that this product is smooth imply that $Ad$ is smooth?
Also in his composition of maps, the first composition
$$g \mapsto \big((g,0), (e,v)\big)$$
is an inclusion map so it is obviously smooth. But why are the other two maps smooth? The second map
$$\big((g,0), (e,v)\big) \rightarrow \big((g,e), (0,v)\big)$$
seems like it should obviously be smooth, but I cannot formulate a proof. The third map,
$$\big((g,e), (0,v)\big) \mapsto D_{(g,e)}c(0,v)$$
is less obvious. I know that the conjugation map $c$ is smooth, but why is its differential smooth (especially in this form)?