As I understand it, the way we interpret the derivative on a curve $\gamma$ on a manifold M $\gamma : \mathbb{R} \rightarrow M$ is through the tangent vector $X_\gamma$. $X_\gamma$ is defined by applying it to a function $f\in C^{\infty}(M)$ via $X_\gamma f = (f \circ\gamma)'(0)$, where we need the function f, since $f\circ\gamma : \mathbb{R} \rightarrow \mathbb{R}$, and thus we can apply the derivative we know from real analysis. It appears that the notation used to denote these tangent vectors is just $\dot{\gamma}(0) := X_\gamma$
However, I'm working on trying to prove that the exponential map is equal to the matrix exponential for the group manifold $G = GL(d, \mathbb{R})$. The curve used is $$\begin{align}\gamma_A : \mathbb{R} &\rightarrow GL(d, \mathbb{R})\\ t&\mapsto e^{tA}\end{align}$$
A property we want to prove is that $\dot{\gamma}(0) = A$. If I treat G like a manifold, than to find $\dot{\gamma}(0)$, I have to take the corresponding tangent vector and act on an arbitrary $f\in C^{\infty}(M)$, so I get something like:
$$\begin{align}X{\gamma_A} f &= (f\circ\gamma_A)'(0)\\ &=\frac{d}{dt}|_{t=0}f(e^{tA})\\ &=f'(e^{tA})|_{t=0} A \end{align}$$
And from here I don't know how to proceed. Obviously we could just take the derivative of $\gamma_A$ in the standard way without having to go through all this stuff with tangent vectors and functions on the manifold, but I'm struggling to see when the definitions of the standard derivative and the derivative on the manifold coincide.