Suppose that $G$ is a Lie group, and let $\mathfrak{g}$ be the Lie algebra of $G$. For each $g \in G$, we get the conjugate map $C_g: G \to G; x \mapsto g x g^{-1}$.
I know that for each $X, Y \in \mathfrak{g}$, that $$[X,Y] = \frac{\partial^2}{\partial s \partial t}\bigg|_{0} C_{\text{exp}(sX)} \left( \text{exp}(tY)\right). \tag{1} \label{eq1}$$ The sketch of a proof is covered here on page 3.
After testing some examples when $ G = \text{GL}_2(\mathbb{R})$ on Mathematica, identifing the Lie algebra of $\text{GL}_2(\mathbb{R})$ with $M_2(\mathbb{R})$, it seems that the identity
$$ [X, Y] = \frac{\partial^2}{\partial s \partial t} \bigg|_{0} \left[\text{exp}(sX), \text{exp}(tY)\right] \tag{2} \label{eq2}$$ may be true for each $X, Y \in \mathfrak{g}$, where the bracket on the right hand side is the commutator in $G$ given by $[g,h] = g h g^{-1} h^{-1}$, $g, h \in G$.
For example, both $(1)$ and $(2)$ are true when $G = \text{GL}_2(\mathbb{R})$, $X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, Y = \begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}. $
Is $(2)$ true in general? If it is, what are some hints to prove it? It is quite similar to the first identity, which at first led me to believe that it is false, but alas it holds for some simple examples.
Using the Baker-Campbell-Hausdorff formula, we can expand the group commutator $[\exp(sX),\exp(tY)]$, ignoring terms of order $\ge 2$ in $s$ or $t$. $$ \exp(sX)\exp(tY)\exp(sX)^{-1}\exp(tY)^{-1} $$ $$ =\exp(sX)\exp(tY)\exp(-sX)\exp(-tY) $$ $$ =\exp\left(sX+tY+\frac{st}{2}[X,Y]+\dots\right)\exp\left(-sX-tY+\frac{st}{2}[X,Y]+\dots\right) $$ $$ =\exp\left(st[X,Y]+\dots\right) $$ This, combined with the appropriate identifications (namely the canonical identification $\mathfrak{g}\cong T_0\mathfrak{g}\cong T_eG$ with the second isomorphism given by $d_0\exp$), gives the desired result.