I'm studying Lie Algebras, following the book "Algebras de Lie - Luiz A. B. San Martin" and page 25 the author says that, the map
$$ \rho:\mathfrak{gl}(n,\mathbb{R}) \to \mathfrak{gl} (\mathcal{C}^{\infty}(\mathbb{R}^n,\mathbb{R}^n)))$$ such that $\rho(A)f(x) = \text{d}f_x(Ax)$ is an example of representation.
I wasn't able to verify that this map is a representation of $\mathfrak{gl}(n,\mathbb{R}) $ on $\mathcal{C}^{\infty}(\mathbb{R}^n,\mathbb{R}^n)$. Can anyone help me?
My Effort
Definition: Let $V$ be a vector space and $\mathfrak{gl}(V)$ the Lie Algebra of the linear transformations on $V$. Moreover, let $\mathfrak{g}$ be a Lie Algebra (over the same field as $V$). A representation of $\mathfrak{g}$ on $V$ is a homomorphism $\rho:\mathfrak{g} \to \mathfrak{gl} (V)$.
Then, I just need to verify that our $\rho$ is a linear transformation between vector spaces and
$$\rho([X,Y]) = [\rho(X),\rho(Y)]. $$
The first part is obvious, once $$\rho(A+\lambda B)f(x) = \text{d}f_x((A+\lambda B)x) =\text{d}f_x(Ax) +\lambda\text{d}f_x(Bx)= (\rho(A)+ \lambda \rho(B) )f(x). $$
My problem is in the verification of $\rho([X,Y]) = [\rho(X),\rho(Y)], $ because $$ \rho([A,B])f = \text{d}f_x((AB - BA)x) =\text{d}f_x(ABx) -\text{d}f_x( BA)x $$
and $[\rho(A),\rho(B)]f(x)$ involves second derivatives, therefore, there can be no equality.
$\rho(B)(\rho(A)(f))=\rho(B)(df_x(A(x))=d^2f_x(B(x).A(x))+df_x(A(B(x))$
since $d^2f_x(B(x).A(x))=d^2f_x(A(x).B(x))$ you deduce the result.