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 $f$.  )

I think the differential at $(e,e)$ is actually zero:
First, if $m : G\times G \to G, (g,h)\mapsto gh$ is the multiplication map then $dm : T_e G \oplus T_e G \to T_e G$ is given by $(X,Y)\mapsto X+Y$.  
Also if $i: G \to G, g \mapsto g^{-1}$ denotes the inversion map then we have $di : T_e G \to T_e G, X\mapsto -X$.
Your map $f$ is given by the composition
$$
G\times G \to G\times G\times G \times G \to G\times G \to G \\
(g,h) \mapsto (g,h, g^{-1}, h^{-1}) \mapsto (gh, g^{-1} h^{-1}) \mapsto ghg^{-1} h^{-1}
$$
so the differential is the composition
$$
(X,Y) \mapsto (X,Y, -X,-Y) \mapsto (X+Y, -X-Y) \mapsto0.
$$
