[proofplan]
A finite-dimensional representation of a Lie group is precisely a smooth Lie [group homomorphism](/page/Group%20Homomorphism) into a general linear group. Therefore its differential at the identity is the [Lie algebra](/page/Lie%20Algebra) map induced by that homomorphism. We apply the general bracket-preservation theorem for differentials of Lie group homomorphisms, then identify the Lie algebra of $GL(V)$ with $\mathfrak{gl}(V)$ and its Lie bracket with the commutator.
[/proofplan]
[step:Regard the representation as a smooth Lie group homomorphism]
By hypothesis,
\begin{align*}
\rho:G\to GL(V)
\end{align*}
is a smooth representation. Thus $\rho$ is a smooth Lie group homomorphism from $G$ to the Lie group $GL(V)$. Its differential at the identity element $e\in G$ is the [linear map](/page/Linear%20Map)
\begin{align*}
d\rho_e:T_eG\to T_{I_V}GL(V),
\end{align*}
where $I_V:V\to V$ denotes the identity linear map.
Under the standard identification
\begin{align*}
T_{I_V}GL(V)\cong \mathfrak{gl}(V)=\operatorname{End}(V),
\end{align*}
the map $d\rho_e$ is the differentiated representation denoted by
\begin{align*}
d\rho:\mathfrak g\to \mathfrak{gl}(V).
\end{align*}
[guided]
The first point is only a matter of unpacking the words in the statement. A finite-dimensional representation of $G$ on $V$ is a smooth Lie group homomorphism
\begin{align*}
\rho:G\to GL(V).
\end{align*}
Since $\rho$ is smooth and sends the identity element $e\in G$ to the identity operator $I_V\in GL(V)$, its differential at $e$ is a well-defined linear map
\begin{align*}
d\rho_e:T_eG\to T_{I_V}GL(V).
\end{align*}
The Lie algebra of $G$ is
\begin{align*}
\mathfrak g:=T_eG.
\end{align*}
The Lie algebra of the general linear group $GL(V)$ is identified with
\begin{align*}
\mathfrak{gl}(V)=\operatorname{End}(V)
\end{align*}
by viewing $GL(V)$ as an open submanifold of the finite-dimensional [vector space](/page/Vector%20Space) $\operatorname{End}(V)$. Under this identification, the differential $d\rho_e$ is exactly the differentiated representation
\begin{align*}
d\rho:\mathfrak g\to \mathfrak{gl}(V).
\end{align*}
[/guided]
[/step]
[step:Apply bracket preservation for differentials of Lie group homomorphisms]
The hypotheses of [citetheorem:8803] are satisfied with $H=GL(V)$ and $\varphi=\rho$: both $G$ and $GL(V)$ are Lie groups, and $\rho:G\to GL(V)$ is a smooth Lie group homomorphism. Hence the differential at the identity
\begin{align*}
d\rho_e:\mathfrak g\to T_{I_V}GL(V)
\end{align*}
preserves Lie brackets. Therefore, for every $X,Y\in\mathfrak g$,
\begin{align*}
d\rho_e([X,Y])=[d\rho_e(X),d\rho_e(Y)]_{\operatorname{Lie}(GL(V))}.
\end{align*}
[guided]
Now we use the structural theorem that makes the proof work: [citetheorem:8803] says that the differential at the identity of a smooth Lie group homomorphism preserves Lie brackets. We must check its hypotheses in the present situation. The source $G$ is a Lie group by hypothesis. The target $GL(V)$ is the general linear group of the finite-dimensional vector space $V$, hence it is a Lie group with identity element $I_V$. Finally, the representation
\begin{align*}
\rho:G\to GL(V)
\end{align*}
is a smooth Lie group homomorphism by the definition of a smooth representation.
Thus [citetheorem:8803] applies with $H=GL(V)$ and $\varphi=\rho$. Its conclusion says that the differential at the identity
\begin{align*}
d\rho_e:\mathfrak g\to T_{I_V}GL(V)
\end{align*}
is a homomorphism of Lie algebras. Therefore, for every $X,Y\in\mathfrak g$,
\begin{align*}
d\rho_e([X,Y])=[d\rho_e(X),d\rho_e(Y)]_{\operatorname{Lie}(GL(V))}.
\end{align*}
Here the bracket on the right is the Lie bracket in the Lie algebra of the target Lie group $GL(V)$. This is not yet the final displayed identity in the theorem statement, because the theorem writes the target Lie algebra as $\mathfrak{gl}(V)$ and writes its bracket as the commutator of endomorphisms. That identification is made in the next step.
[/guided]
[/step]
[step:Identify the target bracket with the commutator bracket]
Under the identification
\begin{align*}
T_{I_V}GL(V)\cong \mathfrak{gl}(V),
\end{align*}
the Lie bracket on the Lie algebra of $GL(V)$ is the commutator bracket on endomorphisms:
\begin{align*}
[A,B]=A\circ B-B\circ A
\end{align*}
for all $A,B\in\mathfrak{gl}(V)$. Substituting $A=d\rho(X)$ and $B=d\rho(Y)$ into the bracket-preservation identity gives
\begin{align*}
d\rho([X,Y])=[d\rho(X),d\rho(Y)].
\end{align*}
Since $X,Y\in\mathfrak g$ were arbitrary, this proves the claim.
[guided]
It remains to translate the bracket-preservation identity from the abstract Lie algebra of $GL(V)$ into the concrete notation used in the theorem statement. Since $V$ is finite-dimensional, $GL(V)$ is an open submanifold of the vector space $\operatorname{End}(V)$ of linear maps $V\to V$. Therefore its tangent space at the identity operator is identified with
\begin{align*}
T_{I_V}GL(V)\cong \operatorname{End}(V)=\mathfrak{gl}(V).
\end{align*}
Under this identification, the Lie bracket on $\operatorname{Lie}(GL(V))$ is the commutator bracket on endomorphisms: for all $A,B\in\mathfrak{gl}(V)$,
\begin{align*}
[A,B]=A\circ B-B\circ A.
\end{align*}
From the previous step, for every $X,Y\in\mathfrak g$ we have
\begin{align*}
d\rho_e([X,Y])=[d\rho_e(X),d\rho_e(Y)]_{\operatorname{Lie}(GL(V))}.
\end{align*}
Using the identification $T_{I_V}GL(V)\cong\mathfrak{gl}(V)$, the map $d\rho_e$ is the differentiated representation
\begin{align*}
d\rho:\mathfrak g\to\mathfrak{gl}(V),
\end{align*}
and the bracket on the right becomes the commutator bracket. Hence
\begin{align*}
d\rho([X,Y])=[d\rho(X),d\rho(Y)].
\end{align*}
Because $X,Y\in\mathfrak g$ were arbitrary, this is exactly the desired identity for all elements of the Lie algebra $\mathfrak g$.
[/guided]
[/step]
