Consider the following fragment from Tu's book An introduction to manifolds (pp.181-182 of the second edition, section 16 "Lie algebra's").
Why exactly is the function
$$F: G \times I \to \mathbb{R}: (g,t) \mapsto \frac{d}{ds}\Big\vert_{s=t} f(g c(t))$$
smooth (= $C^\infty$?) I understand that $(g,s) \mapsto f(gc(s))$ is $C^\infty$, but I don't know why this implies that $F$ is smooth as well, although it is something that can be expected.
Thanks in advance for your help!
This is essentially because of three facts:
The $k$th derivative of a $C^r$ function ($r\geq k$) is $C^{r-k}$, and in particular any derivative of a $C^\infty$ function is $C^\infty$.
Any mixed partial derivative of order $k$ of a $C^r$ function ($r\geq k$) is invariant under permutations of the order of the derivatives; in particular one can take any mixed partial derivative of a $C^\infty$ function in any order one likes.
For $M$, $N$, and $P$ $C^\infty$ manifolds, a function $F:M\times N\to P$ is $C^r$ iff all mixed partials $D_{\text{along }M}^i D_{\text{along }N}^j F$, $i+j\leq r$ exist and are continuous iff all mixed partials $\partial_{x_1}^{i_1}\cdots\partial_{x_m}^{i_m}\partial_{y_1}^{j_1}\cdots\partial_{y_n}^{j_n} F$, $i_1+\cdots + i_m+ j_1 +\cdots +j_n\leq r$ exist and are continuous, where $(x_1,...,x_m)$ and $(y_1,...,y_n)$ are anonymous local charts for $M$ and $N$, respectively. (And $F$ is $C^\infty$ iff it's $C^r$ for any $r\in\mathbb{Z}_{\geq1}$.)
Thus since $(g,t)\mapsto f(gc(t))$ is $C^\infty$ one has
$$D^i_{\text{along }G}D^j_{\text{along }I}F(g,t) =D^i_{\text{along }G}D^j_{\text{along }I}\dfrac{d}{dt}f(gc(t)) =D^i_{\text{along }G}D^{j+1}_{\text{along }I}f(gc(t)),$$
whence $F$ is $C^\infty$.
(Or: $D^i_{\text{along }G}D^j_{\text{along }I}F(g,t) =\dfrac{d}{dt} D^i_{\text{along }G}D^j_{\text{along }I}f(gc(t))$)
It might be instructive to consider the case of $G=\mathbb{R}$.