The diffential of commutator map in a Lie group Leb $G$ be a Lie group and $f:G\times G\rightarrow G$ be the commutator map $:(x,y)\mapsto xyx^{-1}y^{-1}$.
How to obtain the Lie bracket in the associated Lie algebra of $G$ from the derivatives of $f$?
(We know that the Lie bracket is defined via the adjoint representation.)
(I saw $df_{(e,e)}(X,Y)=[X,Y]$ somewhere. Thanks to @John for remarking this is false. But I think there is indeed a relation between the Lie bracket and derivatives of $f$.  )

Let $G$ be a Lie group with the neutral element $e$. Then, given two elements in its Lie algebra, $\xi,\eta\in\operatorname{Lie} G:=T_eG$, represented by (smooth) paths $x,y\colon (-\varepsilon,\varepsilon)\to G$ with $x(0)=y(0)=e$, one can define the Lie bracket as the second derivative of the commutator map: $$[\xi,\eta]=\frac{d^2}{dt^2}(x(t)y(t)x(t)^{-1}y(t)^{-1})\biggr|_{t=0}.$$
Note that the first derivative of the commutator vanishes, which makes the second derivative well-defined as an element of the tangent space $T_eG$.
