 Solution

First simplify the composite: $(S\circ \mathcal{I}\circ S)(A)=((A^2{)}^{-1}{)}^2=A^{-4}$. Applying the product rule to four factors of $A^{-1}$, and the preceding inverse derivative to one factor at a time, givesThe four terms have left and right exponents $(1,4),(2,3),(3,2),(4,1)$. The same expression follows by composing $D{S}_{A^{-2}}$, $D{\mathcal{I}}_{A^2}$ and $D{S}_A$ in that order. Only when $\Delta$ commutes with $A$ can it be simplified to $-4A^{-5}\Delta$.