Given any matrix $A$, one wishes to give an expansion at $A$, and especially wishes to unravel a possible differentiation of the $\\\exp: Mat(n, \mathbb{C}) \rightarrow GL(n, \mathbb{C})$ mapping. Now, the difficulty arises as we don't have any binomial theorem for non-commutative entities.
$\\\exp(A+H) = Id + (A+H) + \frac{1}{2}(A+H)^2 + \frac{1}{6}(A+H)^3 + ... $
Let $\Lambda_k(H) = \sum_{i+j= k-1} A^iHA^j$. Then $\Lambda_k$ represents all terms linear in $H$ in $(A+H)^k$. We then define $\Lambda = \sum_{k} \frac{1}{k!}\Lambda_k$. By introducing operators $L_A$ and $R_A$ such that $L_A(H) = AH$ and $R_A(H) = HA$, we can rewrite $\Lambda = \sum_{k} \frac{1}{k!}\sum_{i+j=k-1} L_A^iR_A^j$. As $L_A$ and $R_A$ commute, by doing formal calculus using $\frac{x^n - y^n}{x-y} = \sum_{i+j = k-1} x^iy^j$, it yields $exp'(X) = L_{\\\exp(X)}\frac{1-e^{-[X,.]}}{[X,.]}$. Moreover, one can easily show that indeed:
$\\\exp(A+H)=$ $\\\exp(A) + \Lambda(H) + O(||H||^2)$
Now, I was wondering : can one give an explicit formula for the $O(||H||^2)$?
Actually, there is the Zassenhaus formula, expressed unsurprisingly in terms of commutators.
I would say that counting the number of commutators you will get the right bound.