Derivative of the matrix exponential ($\frac{d}{dt}{e}^{f(t)}={\int }_0^1{e}^{(1-u)f(t)}{f}^{\prime }(t)e^{uf(t)}du$)

This formula allows a matrix and its derivative to fail to commute. Differentiate the absolutely convergent power series of the matrix exponential to obtain ${\sum }_{k\ge 1}k{!}^{-1}{\sum }_{j=0}^{k-1}{f}^j{f}^{\prime }{f}^{k-1-j}$. Expanding the integral gives the same coefficients, since ${\int }_0^1(1-u{)}^a{u}^{b}du=a!b!/(a+b+1)!$. The formula reduces to $e^f{f}^{\prime }$ when $[f,f^{\prime }]=0$.