Proof that the adjoint representation of a Lie group is linear The definition of the adjoint representation ad of a Lie group $G$ that I've been looking at is: $$\text{for } g\in G,X\in\mathfrak{g},\ \text{ad}_gX=\frac{d}{dt}(ge^{tX}g^{-1})|_{t=0}.$$ I want to prove that $\text{ad}_g(X+Y)=\text{ad}_g(X)+\text{ad}_g(Y)$ in the general context of any Lie group, not just for matrix Lie groups (scalar multiplication is straightforward). 
Is the following on the right track? $$\text{ad}_g(X+Y)=\frac{d}{dt}(ge^{t(X+Y)}g^{-1})|_{t=0}=\frac{d}{dt}(g\gamma^{X}(t)\gamma^{Y}(t)\gamma^{-\frac{1}{2}[X,Y]}(t^2)\cdots g^{-1})|_{t=0}=?$$ where $\gamma^{X}(t)$ is the integral curve of the left-invariant field associated with $X$ with initial condition $\gamma^{X}(0)=e$. 
Thank you very much for your help!
 A: For a Lie group G, the map $\text{Ad}_g:G\rightarrow G$ that takes $h\in G$ to $ghg^{-1}$ is smooth and invertible, and its inverse is also smooth. Laying down a coordinate chart $\varphi:g\mapsto(x^1,\cdots,x^n)$ in a neighbourhood of the identity element $e$, we can define functions from old to new coordinates under $\text{Ad}_g$, namely $x'^i:=x^i\circ\text{Ad}_g\circ\varphi^{-1}=x'^i(x^j)$. The properties of $\text{Ad}_g$ means that $\frac{\partial x'^i}{\partial x^j}$ is well-defined and non-singular.
The adjoint representation is then given, component-wise, by $[\text{ad}_g(X)]^i=[\frac{d}{dt}\text{Ad}_g(e^{tX})|_{t=0}]^i=\frac{dx'^i(x^j(\gamma(t)))}{dt}|_{t=0}$, where $\gamma(t)$ is the integral curve through $e$ with tangent vector $X$. By the chain rule, this equals $\frac{\partial x'^i}{\partial x^j}\frac{dx^j(\gamma(t))}{dt}|_{t=0}=\frac{\partial x'^i}{\partial x^j}X^j$. Therefore the representation is clearly linear and is given by the matrix $\frac{\partial x'^i}{\partial x^j}$. 
A: From the comments:
Since conjugation by $g$ is a smooth map taking $e$ to $e$, its derivative (at $e$) is a linear map from $T_eG$ to $T_eG$. – Max 22 hours ago
