I found the following on my notes of differential geometry:
Proposition: Let $G$ be a Lie group with Lie algebra $\mathfrak{g}=T_1G \cong \mathbb{R}^n$. The exponential map $\text{exp} \colon \mathfrak{g} \to G$ is smooth and $D_0\text{exp} \colon \mathfrak{g} \to \mathfrak{g}$ is the identity.
Proof: Identifying $T_0\mathfrak{g} \cong \mathfrak{g}$ we get $D_0\text{exp}(x) = \frac{d}{dt} \mid _{t=0}\text{exp}(tx)=x.$
My first question is: is it so obvious that exp is smooth that it is not even worth mentioning the proof, or is it just that one should check it with the charts and it's boring?
My second question is: $t \to \text{exp}(tx)$ is a function $\mathbb{R} \to G$, so what does the derivative $\frac{d}{dt}$ exactly mean? Why is the derivative at $t=0$ equal to $x$?