Let $G$ be a real Lie group, $V$ a finite dimensional complex vector space, and $\pi: G \rightarrow \operatorname{GL}(V)$ a continuous (hence smooth) representation of $G$. Let $W$ be a linear subspace of $V$. I'm trying to understand why if $W$ is stable under $G$, then $W$ is also stable under $\mathfrak g = \operatorname{Lie} G$ via the tangent map $d\pi: \mathfrak g \rightarrow \operatorname{End}(V)$.
If $X \in \mathfrak g$, then the exponential map defines a Lie group homomorphism $\alpha(t) = \exp(tX)$ from $\mathbb R$ to $G$ from which we recover $X$ as $\alpha'(0)$ (here we are identifying $\alpha'$ as a smooth map from $\mathbb R$ to the tangent bundle $T(G)$).
We want to show that if $v \in W$, then so is $d\pi(X)v$. By hypothesis, $\pi(\alpha(t))v \in W$ for every $t \in \mathbb R$. By the chain rule, the derivative of $t \mapsto \pi \circ \alpha(t)$ is the composition of the derivatives, i.e. $d\pi(\alpha'(t))$. Plugging in $t = 0$, we get
$$\frac{d}{dt}|_{t=0} \pi(\alpha(t)) = d\pi(X)$$
And therefore
$$[\frac{d}{dt}|_{t=0} \pi(\alpha(t))]v = d\pi(X)v \tag{$\ast$}$$
I would be done if I could interpret the left hand side as something like a classical derivative:
$$\lim\limits_{h\to 0} \frac{1}{h} (\pi(\alpha(h))v - v)$$
But taking the derivative of $\pi(\alpha(t)))$ at any point $h$ near $0$ will produce only a tangent vector at a tangent space near $\mathfrak g$. Is such an interpretation of the left hand side of $\ast$ possible?
Such an interpretation is possible. The tangent space map (derivative) is defined abstractly as a collection of linear transformations on different tangent spaces, but the choice of charts lets us locally consider the tangent space map as an ordinary derivative.
Choose a basis for $V$ (considered as a real vector space), so $V$ identifies with $\mathbb R^{2n}$. Fix $v \in V$. Identifying the tangent space of $V$ with itself, the derivative (tangent space map) at $1_G$ of the smooth map $G \rightarrow V$ sending $g$ to $\pi(g)v$ is $X \mapsto d \pi(X)v$. Therefore by the chain rule, the derivative at $0$ of the smooth map $\beta: \mathbb R \rightarrow \mathbb R^{2n}$ sending $t$ to $\pi(\alpha(t))v$ is $d\pi(\alpha'(t))v$. That is,
$$d\pi(X)v = \frac{d}{dt}|_{t=0}[\pi(\alpha(t))v] = \beta'(0)$$
The choice of the chart for $V$ allows us to simultaneously identify different tangent spaces of $V$ together, i.e. we are in the situation where $\beta: \mathbb R \rightarrow \mathbb R^{2n}$ acts like an ordinary derivative. So we really can say
$$\beta'(0) = \lim\limits_{h\to 0} \frac{1}{h}(\beta(h) - \beta(0)) = \lim\limits_{h \to 0} \frac{1}{h}(\pi(\alpha(t))v - v) \in W$$