Derivative of the matrix exponential
= Derivative of the matrix exponential
{title2=$\frac{d}{dt}e^{f(t)}=\int_0^1e^{(1-u)f(t)}f'(t)e^{uf(t)}\,du$}
This formula allows a <matrix> and its derivative to fail to commute. Differentiate the absolutely convergent <power series> of the <matrix exponential> to obtain $\sum_{k\geq1}k!^{-1}\sum_{j=0}^{k-1}f^jf'f^{k-1-j}$. Expanding the integral gives the same coefficients, since $\int_0^1(1-u)^au^b\,du=a!b!/(a+b+1)!$. The formula reduces to $e^ff'$ when $[f,f']=0$.