高维微分学—向量值映照的可微性 谢锡麟 3.设A(t)∈Rmxn,B(t)∈Rnx,有 Rt→(AB)(t)=A(t)B(t)∈R 则有 dB AB (t)B(t)+A(t)-(t) 设A(t)∈Rmxn,B(1)∈Rnx,C(t)∈R×P,有 R3t(ABC)(t)=A(t)B(t)C(t)∈Rn×P 则有 d dA dB dc (ABC)(t)=(t)B(C(t)+A(t)()C()+A(t)B(t)x(t)∈R"× 证明1.考虑 A in (t) a 则有 (Ab)()=∑Ab(t) 所以 (Abi(t) ∑ 因此,有 db (Ab)(t)=A(t)∈R 2.由(Ab)(t)=∑A3(t)b(t),有 da d bs b()=∑(()+∑4()( 即有 dA Ab(t=(t)b(t)+A(t)-(t) 3.由 (AB)()=∑A(t)B8(t,1≤im,1≤j≤1 则有 dB AB)ii (t) (t)Bk; (t)+> Aik(微积分讲稿 谢锡麟 高维微分学—— 向量值映照的可微性 谢锡麟 3. 设 A(t) ∈ R m×n，B(t) ∈ R n×l , 有 R ∋ t 7→ (AB)(t) = A(t)B(t) ∈ R m×l , 则有 d dt (AB)(t) = dA dt (t)B(t) + A(t) dB dt (t). 4. 设 A(t) ∈ R m×n , B(t) ∈ R n×l , C(t) ∈ R l×p , 有 R ∋ t 7→ (ABC)(t) = A(t)B(t)C(t) ∈ R m×p , 则有 d dt (ABC)(t) = dA dt (t)B(t)C(t) + A(t) dB dt (t)C(t) + A(t)B(t) dC dt (t) ∈ R m×p . 证明 1. 考虑 Ab(t) =           A11 · · · A1n . . . . . . Ai1 · · · Ain . . . . . . Am1 · · · Amn               b 1 . . . b n     (t) ∈ R m 则有 (Ab)i(t) = ∑n s=1 Aisb s (t) 所以 d dt (Ab)i(t) = ∑n s=1 Ais db s dt (t). 因此，有 d dt (Ab)(t) = A db dt (t) ∈ R n . 2. 由 (Ab)i(t) = ∑n s=1 Ais(t)b s (t), 有 d dt (Ab)i(t) = ∑n s=1 dAis dt (t)b s (t) +∑n s=1 Ais(t) d, bs dt (t) 即有 d dt (Ab)(t) = dA dt (t)b(t) + A(t) db dt (t) ∈ R n . 3. 由 (AB)ij (t) = ∑n k=1 Aik(t)Bkj (t), 1 6 i 6 m, 1 6 j 6 l 则有 d dt (AB)ij (t) = ∑n k=1 dAik dt (t)Bkj (t) +∑n k=1 Aik(t) dBkj dt (t), 10