How to prove the Lie bracket is infinitesimal commutator I am currently studying Lie groups and I cannot solve the following exercise, which I think is vital to my understanding.
The Lie bracket is defined as $[X,Y]=\text{ad}(X)Y$. Let the group commutator be defined by $c(x,y)=xyx^{-1}y^{-1}$. Show that
$$[X,Y]=\frac{d}{ds}\frac{d}{dt}c(\exp sX,\exp tY) \text{ at } s=0,t=0.$$
If I write it out, I get
$$[X,Y]=\text{ad}(X)(Y)=\frac{d}{ds}\text{Ad}(\exp sX)(Y)=\frac{d}{ds}\frac{d}{dt}C_{\exp sX}\exp tY=\frac{d}{ds}\frac{d}{dt}\exp sX \exp tY (\exp sX)^{-1}$$
What am I doing wrong?
 A: The formula is correct. Your approach will work and it should become clear what to do after  putting in a cleverly placed "$-0$", namely consider that 
$$\frac{d}{ds}\text{Ad}(\exp sX)(Y)|_{s=0}=\frac{d}{ds}(\text{Ad}(\exp sX)(Y)-Y)|_{s=0}$$
Now use the fact that 
$$\frac{d}{dt}f(t,t)|_{t=0}=\frac{d}{dt}f(t,0)|_{t=0}+\frac{d}{dt}f(0,t)|_{t=0}$$
for every smooth map $f:U\rightarrow M$ from a neigborhood $U$ of $0$ in $\mathbb{R}^2$ to 
a smooth manifold $M$. 
Here it is shown that the statement $d_{(e,e)}c(X,Y)=[X,Y]$ is false. Note however that $$d_{(e,e)}c(X,Y)=\frac{d}{dt}\exp(tX)\exp(tY)\exp(-tX)\exp(-tY)|_{t=0}$$ which is not the same as the formula you wrote. I think a source for what you are after is given in the comments at that post though.  
Naming the Lie group $G$, the difference comes from either viewing $c$ as a smooth map from $G\times G$  to $G$ or as a smooth map from $G$ to $G$ parametrized by $G$. The latter is what you want and it yields first (differentiation w.r.t. $t$) a smooth map from $G$ to $End(T_eG)$ given by 
$$x\mapsto Ad(x)-Id$$ and then 
(differentiation w.r.t. $s$) a linear map from $T_eG$ to $T_0End(T_eG)=End(T_eG)$ given by 
$$X\mapsto ad(X).$$ 
