Let $f : \Bbb R^{n\times n} \to \Bbb R^{n \times n}$ be the matrix exponential $$ f(A) = \sum_{k = 0}^\infty \frac{A^k}{k!} $$ Let $B$ be a matrix that commutes with $A$. Show that the value of Fréchet derivative $J_{f,A}: \Bbb R \to \Bbb R$ of $f$ at arbitrary point $A$, evaluated at $B$ is given by $$ J_{f,A}(B) = Be^{A} $$
If $B$ does not commute with $A$, the Fréchet derivative is $$ \int_0^1 {e^{(1-s)A}B{e^{sA}}} \,{\rm d}s $$ How can I prove it?