= Solution
{c}
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, gives
$$
\boxed{D(S\circ\mathcal I\circ S)_A(\Delta)
=-\sum_{j=1}^4 A^{-j}\Delta A^{-(5-j)}.}
$$
The four terms have left and right exponents $(1,4),(2,3),(3,2),(4,1)$. The same expression follows by composing $DS_{A^{-2}}$, $D\mathcal I_{A^2}$ and $DS_A$ in that order. Only when $\Delta$ commutes with $A$ can it be simplified to $-4A^{-5}\Delta$.
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