Here's the setup: I have $SL(2;\mathbb{C})$ acting on $V = \mathbb{C}[z,w] = \oplus_d V_d$, where $V_d$ is the homogeneous complex polynomials of degree $d$. The action is precomposition: $\pi(g)f(z,w) = f(g^{-1}(z,w))$. This is a representation with $(\pi,V_d)$ an irreducible subrepresentation for each $d$.
As a representation, it induces an $\mathfrak{sl}(2;\mathbb{C})$ action on $V_d$ via $d\pi_\mathbb{1}:\mathfrak{sl}(2;\mathbb{C})\to \operatorname{End}(V)$. I want to compute $d\pi$ in coordinates so I can explicitly write down the action of the generators $H,E$, and $F$ on each $V_d$.
Naively, I would do this: choose $z^kw^{d-k}$ as a basis of $V_d$, compute $\pi\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \operatorname{Aut}(V_d)\cong GL(d+1;\mathbb{C}),$ differentiate each coordinate function of $\pi$ in each direction $\mathbb{C}^4$, and evaluate at the identity.
But I'm worried about this strategy. Since $SL(2;\mathbb{C})$ is a submanifold of $GL(2;\mathbb{C})$ defined by $\ker(\det)$, $\partial_a$, $\partial_b$, $\partial_c$, and $\partial_d$ are not tangent to $SL(2;\mathbb{C})$. I don't see any reason this will actually give me the $\mathfrak{sl}(2;\mathbb{C})$ action.
So, question:
Will this naive strategy work for reasons I don't see, or should I choose some other coordinate system about the identity --- say, $e^H,e^E,e^F$?
More generally,
If $N\subset M$ is given by $N = F^{-1}(0)$, where $F:M\to X$ is a smooth map and $0\in X$ is a regular value, and $f:\nu N\to Y$ is a smooth map defined on a neighborhood of $N$ in $M$, can one compute $d(f|_N)$ by choosing a coordinate system on $M$ which does not necessarily restrict to coordinates on $N$ and computing $df$?
I'm not sure why I'm confused today about this, but I am. (Perhaps I need more coffee.) I'd be grateful for some clarity. (Also, as an aside, if I'm wrong about any of the representation theory I outlined or, god forbid, the smooth manifold theory, I'd appreciate correction.)
By the way, this is from a homework assignment, so please DO NOT answer with the explicit $\mathfrak{sl}(2;\mathbb{C})$ action!