Past exam of the options I course of the University of Oxford / 2004 / F / F1

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

Expand at a matrix increment $H$:The last term is $O(\Vert H{\Vert }^2)$ in a matrix norm, so it contributes no derivative. HenceThis is a linear operator on matrix increments, not a scalar entrywise derivative.

The requested base point is the identity matrix $I$, not the zero matrix. Since $I$ commutes with every $H$,The matrix power series justifies the quadratic remainder near zero. ConsequentlyAt zero the derivative would instead be $H$; replacing the printed base point would lose the scalar factor.

Here $\exp (I)=eI$. Applying part (a) at that matrix gives $D{s}_{eI}(K)=e(K^t+K)$, while part (b) gives $D{\exp }_I(H)=eH$. ThereforeBoth the outer evaluation point and the inner derivative contribute a factor of $e$.