I'm trying to check that $\rho\colon{\frak gl}(n, \Bbb R) \to {\frak gl}({\cal C}^\infty(\Bbb R^n, \Bbb R^n))$, given by $\rho(A)\colon {\cal C}^\infty(\Bbb R^n, \Bbb R^n)\to {\cal C}^\infty(\Bbb R^n, \Bbb R^n)$, where $\rho(A)(f)\colon \Bbb R^n \to \Bbb R^n$ acts by $\rho(A)(f)(x) = {\rm d}f_x(Ax)$ is a representation.
I managed to check that $\rho$ is linear, but I'm having trouble with $[\rho(A),\rho(B)] = \rho([A,B])$.
If I'm not messing up here too bad, I got that: $$[\rho(A),\rho(B)](f)(x) = {\rm d}(\rho(B)(f))_x(Ax) - {\rm d}(\rho(A)(f))_x(Bx)$$and $$\rho([A,B])(f)(x) = {\rm d}f_x(ABx) - {\rm d}f_x(BAx)$$ but now I'm stuck. Since $\rho(A)(f),\rho(B)(f)\in {\cal C}^\infty(\Bbb R^n, \Bbb R^n)$, it's legitimate to take their derivatives, but I'm not sure of how to do this. Fixed $x$, ${\rm d}f_x$ is linear, but $x \mapsto {\rm d}f_x$ isn't. And I'm not seeing how should the Hessians that appear cancel with each other, if it comes to that.
Help?
Edit: So far: $$\rho([A,B])(f)(x) = \begin{bmatrix} \sum_{j,k,\ell=1}^n(a_{\ell k}b_{kj}-b_{\ell k}a_{kj})\frac{\partial f_1}{\partial x_\ell}(x) x_j\\ \vdots \\ \sum_{j,k,\ell=1}^n(a_{\ell k}b_{kj}-b_{\ell k}a_{kj})\frac{\partial f_n}{\partial x_\ell}(x) x_j\end{bmatrix},$$
also: $${\rm d}(\rho(A)(f))_x = \begin{bmatrix} \sum_{j,k=1}^n a_{kj}\frac{\partial^2f_\ell}{\partial x_m\partial x_k}(x)x_j + \sum_{k=1}^n a_{km}\frac{\partial f_\ell}{\partial x_k}(x)\end{bmatrix}_{1 \leq \ell, m \leq n}$$and with this: $$ {\rm d}(\rho(B)(f))_x(Ax) = \begin{bmatrix} \sum_{j,k,\ell,m=1}^n b_{kj}a_{m\ell}x_jx_\ell \frac{\partial^2 f_1}{\partial x_m \partial x_k}(x) + \sum_{j,k,\ell=1}^n b_{km}a_{m\ell}x_\ell \frac{\partial f_1}{\partial x_k}(x)\\ \vdots \\ \sum_{j,k,\ell,m=1}^n b_{kj}a_{m\ell}x_jx_\ell \frac{\partial^2 f_n}{\partial x_m \partial x_k}(x) + \sum_{j,k,\ell=1}^n b_{km}a_{m\ell}x_\ell \frac{\partial f_n}{\partial x_k}(x)\end{bmatrix}$$
Essentialy to compute the expression we want, we swap $a \leftrightarrow b$ above and subtract, so after renaming: $$ {\rm d}(\rho(B))(f)_x(Ax) - {\rm d}(\rho(A)(f))_x(Bx) = \begin{bmatrix} \sum_{j,k,\ell=1}^n(a_{kj}b_{k\ell}-a_{\ell k}b_{kj})\frac{\partial f_1}{\partial x_\ell}(x)x_j \\ \vdots \\ \sum_{j,k,\ell=1}^n(a_{kj}b_{k\ell}-a_{\ell k}b_{kj})\frac{\partial f_n}{\partial x_\ell}(x)x_j \end{bmatrix}$$
I'm off by a sign. What is wrong? And I'm still interested in a non-medieval solution to this...