Second Order Taylor Expansion of $\exp^{-1}(\exp tX \exp tY)$ I'm attempting to prove that the product of exponentials $\exp tX \exp tY$ in a Lie group can be expanded to second order as
$$\exp^{-1}(\exp tX \exp tY) = t(X+Y) + \frac12 t^2[X,Y] + t^3\widehat{Z}(t).$$
I've found some lecture notes here which derive the expansion using the fact that
$$(Xf)(a) = \frac{d}{dt}\bigg|_{t=0} f(a\exp tX),$$
where $f$ is a smooth function on a Lie group $G$ and $X$ is a left invariant vector field on $G$. These notes are almost perfectly clear to me, aside from one small piece of connective tissue. When forming the Taylor sum, the notes claim in a proposition
$$f(\exp tX_1 \dots \exp tX_n) = f(e) + t\sum_i X_if(e) + \frac{t^2}{2}\left\{\sum_i X_i^2f(e) + 2\sum_{i<j}X_iX_jf(e) \right\} + O(t^3),$$
which would imply that, for the first order term,
$$\frac{d}{dt}\bigg|_{t=0} f(\exp tX_1 \dots \exp tX_n) = \sum_i X_if(e) = \sum_i \frac{d}{dt}\bigg|_{t=0} f(\exp tX_i).$$
Unfortunately, I can't manage to convince myself that the left and right sides of the above are, in fact, equal. However, I'm confident that once I've worked out the issue there, the second order term will follow immediately.
What obvious fact am I missing here? Exactly how dense am I being?
 A: Thanks to the comments left by Ted Shifrin, I believe I've ironed out the problem! The "trick" seems to be defining a set of ostensibly independent variables $t_1,\dots, t_n$ which are all, in fact, equal to $t$. With this, the function in question becomes $f(\exp t_1X_1 \dots \exp t_nX_n)$, and I can then apply the chain rule:
$$
\frac{d}{dt}\bigg|_{t=0}f(\exp t_1X_1 \dots \exp t_nX_n) = \sum_i\frac{\partial}{\partial t_i}\bigg|_{t_1=\dots=t_n=0} f(\exp t_1X_1 \dots \exp t_nX_n) \frac{d}{dt}t_i.
$$
But of course, by the definition of the variables $t_i$, it's clear that $\frac{d}{dt}t_i = 1$. Thus, the derivative becomes
$$\begin{align}
\frac{d}{dt}\bigg|_{t=0}f(\exp t_1X_1 \dots \exp t_nX_n) &= \sum_i\frac{\partial}{\partial t_i}\bigg|_{t_1=\dots=t_n=0} f(\exp t_1X_1 \dots \exp t_nX_n) \\
&= \sum_i \frac{\partial}{\partial t_i}\bigg|_{t_1=\dots=t_n=0}(\exp t_1X_1 \dots \exp t_nX_n)f \\
&= \sum_i \exp' t_iX_i|_{t_i=0}f \\
&= \sum_i X_i|_ef = \sum_i (X_if)(e),
\end{align}$$
where on the second line I've interpreted $\exp t_1X_1 \dots \exp t_nX_n = \gamma(t_i)$ as a path on $G$, and used the fact that $(f \circ \gamma)'(t_0) = \gamma'(t_0)f.$ Thank you Ted Shifrin for the tip!
