故det(I+xA) = (1+λ1·x)(1+λ2·x)...(1+λm·x),而tr(A(I+xA)^(-1)) = λ1/(1+λ1·x)+λ2/(1+λ2·x)+...+λm/(1+λm·x).于是(ln(det(I+xA)))' = (ln((1+λ1·x)(1+λ2·x)...(1+λm·x)))'= (ln(1+λ1·x))'+(ln(1+λ2·x...