Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 2 / c

At a fixed $t$, write $A=f(t)$ and $B=f^{\prime }(t)$. Differentiating the matrix exponential in the direction $B$ givesThe order of factors matters because matrix multiplication need not be commutative.

To prove the derivative of the matrix exponential formula, expand its power series. The coefficient linear in $h$ in $(A+hB{)}^k$ is ${\sum }_{j=0}^{k-1}{A}^{j}B{A}^{k-1-j}$. On bounded sets of matrices, the exponential series and its directional derivative series converge uniformly: the latter is dominated in operator norm by $\Vert B\Vert {\sum }_{k\ge 1}k\Vert A{\Vert }^{k-1}/k!$. Terms with two or more $hB$ factors have sum $O(h^2)$ locally. Hence the derivative isOn the other hand, the integral in the box expands absolutely asThe integral identity follows by repeated integration by parts. Grouping by $a+b=k-1$ gives the same derivative. Finally differentiability means $f(t+h)=A+hB+o(h)$; the locally bounded derivative of the exponential sends the $o(h)$ error to $o(h)$, proving the chain rule even when $f^{\prime }$ is not continuous. If $[f,f^{\prime }]=0$, the formula simplifies to $e^f{f}^{\prime }$.