Let $G$ be a linear algebraic group over an algebraically closed field of characteristic zero. Let $X$ be an affine variety on which $G$ acts. Then $G$ naturally acts on the coordinate ring $\mathcal O_X(X)$ by the formula $g.f(x) = f(g^{-1}.x)$. I often see the claim "The Lie algebra $\mathfrak g$ of $G$ also acts on $\mathcal O_X(X)$ by derivations." What exactly does this mean?
For example, suppose we take $X = G$, with $G$ acting by conjugation. If $\xi \in \mathfrak g$, and $f$ is a regular function on $G$, what would $\xi.f$ be?
I'm a little more familiar with the case of real Lie groups. If $G$ was a smooth Lie group with an action on a smooth manifold $X$, and $f$ was a smooth real valued function on $X$, then I would expect $\xi.f$ would be
$$\xi.f(x) = \lim\limits_{t \to 0} \frac{f(\exp(t \xi).x )-f(x)}{t}$$
By definition, an element of $\mathfrak{g}$ is a tangent vector of $G$ at the identity, which we can think of as a derivation $\xi:\mathcal{O}_G(G)\to k$ where $k$ is the base field. Now the action morphism $\mu:G\times X\to X$ gives a $k$-algebra homomorphism $\mu^*:\mathcal{O}_X(X)\to \mathcal{O}_G(G)\otimes\mathcal{O}_X(X)$. We can then compose this map with $\xi\otimes 1_{\mathcal{O}_X(X)}:\mathcal{O}_G(G)\otimes\mathcal{O}_X(X)\to\mathcal{O}_X(X)$ to get a $k$-linear map $\mathcal{O}_X(X)\to\mathcal{O}_X(X)$ which is a derivation since $\mu^*$ is an algebra homomorphism and $\xi$ is a derivation.
(It is a good exercise to check that in the case of real Lie groups this coincides with the analytic definition. The key point is that $\xi$ acts as a derivation on $\mathcal{O}_G(G)$ exactly by taking a derivative along the curve $\exp(t\xi)$.)