From the Wikipedia page of the Adjoint Representation we can read
If $G$ is an immersed Lie subgroup of the general linear group $\mathrm {GL} _{n}(\mathbb {C} )$ (called immersely linear Lie group), then the Lie algebra ${\mathfrak {g}}$ consists of matrices and the exponential map is the matrix exponential $\operatorname {exp} (X)=e^{X}$ for matrices $X$ with small operator norms. We will compute the derivative of $\Psi _{g}$ at $e$. For $g$ in $G$ and small $X$ in ${\mathfrak {g}}$, the curve ${\displaystyle t\to \exp(tX)}$ has derivative $X$ at $t = 0$, one then gets: $${\displaystyle \operatorname {Ad} _{g}(X)=(d\Psi _{g})_{e}(X)=(\psi _{g}\circ \exp(tX))'(0)=(g\exp(tX)g^{-1})'(0)=gXg^{-1}}$$ where on the right we have the products of matrices. If ${\displaystyle G\subset \mathrm {GL} _{n}(\mathbb {C} )}$ is a closed subgroup (that is, $G$ is a matrix Lie group), then this formula is valid for all g in G and all $X$ in ${\mathfrak {g}}$.
My question is: Why is it necessary for $X$ to be an small element in ${\mathfrak {g}}$?
I don't see this being used at any point of the proof, and I believe that the exponential map should be well defined for matrices as the usual Taylor series.
Is the smallness of the group element necessary to ensure differentiability at the origin? I don't se why, as we're differentiating with respect to $t$, and not to $X$.
Many thanks!