Question: ah! based on that definition, if g(t) = W(t) * e^[(from t_0 to t) tr(A(s)) ds] then g'(t) = [W'(t)-W(t)trA(t)] * e^[(from t_0 to t)
ah! based on that definition, if g(t) = W(t) * e^[(from t_0 to t) tr(A(s)) ds] then g'(t) = [W'(t)-W(t)trA(t)] * e^[(from t_0 to t) tr(A(s)) ds] which would then = 0 since W'(t) = tr(A(t)) * W(t)
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