解z= f(u, v,w) = f(u,p(u,s),w) = f(u,p(u, (u,w),w)由存规则,0zafauafafaf000s(04ou+auJuduJududuasduaf .afafaaap=Ju+JuauJuauasazafafauafaf(a00safaf=dwOwOdwOuCoswdwduasow最后，我定简单地为绍全微其（形式微其)的概于.果f：D→R为超微的多元函数，由定义，f改处的微其df()是一个线性映射Rm-Rdf(r) :2%V我定把映射→df(r）称为f的全微其，记为df.由于d(f +μg)(r) = Adf(r) + μdg(r), A,μE R,因例，全微其之间也超以定义它法和数来运算，改这个意义下，有df =(*)d(f +μg)=Adf +μdg,d(f,g) = f-dg+gdf,gdf - fdgd(f) =(g + 0).g29如果将(*)写成下阵形式df = Jf(da1,...,dan)T则复合映射的存规则超写为d(fog)=J(fog)-(dui,..,dum)T,=Jf(r).Jg(u)(du1,***,dum)T(r= g(u))=Jf(d(u))- (dg1,**,dgn)T11
z = f(u,v,w) = f(u,ϕ(u,s),w) = f(u,ϕ(u,ψ(u,w)),w). >&_N, ∂z ∂u = ∂f ∂u + ∂f ∂v · ∂v ∂u = ∂f ∂u + ∂f ∂v · ∂ϕ ∂u + ∂ϕ ∂s · ∂s ∂u = ∂f ∂u + ∂f ∂v · ∂ϕ ∂u + ∂f ∂v · ∂ϕ ∂s · ∂ψ ∂u ∂z ∂w = ∂f ∂w + ∂f ∂v · ∂v ∂w = ∂f ∂w + ∂f ∂v ∂ϕ ∂s · ∂s ∂w = ∂f ∂w + ∂f ∂v · ∂ϕ ∂s · ∂ψ ∂w. lk, Æ7}*3 _SG ( gG) 1NB. a f : D → R 1 >Gdp, >76, f M x "1G df(x) j/Q!<` df(x) : R m → R v 7→ Xm i=1 ∂f ∂xi · vi Æ7<` x 7→ df(x) f 1SG, x df. >B d(λf + µg)(x) = λdf(x) + µdg(x), λ,µ ∈ R, 7#, SGX|.276yAgp Ku, MRQ46, ? df = Xn i=1 ∂f ∂xi · dxi (∗) d(λf + µg) = λdf + µdg, d(f,g) = f · dg + gdf, d( f g ) = gdf − fdg g 2 (g 6= 0). Za (*) S g: df = Jf(dx1, · · · ,dxn) T NIi<`1&_N d(f ◦ g) = J(f ◦ g) · (du1, · · · ,dum) T , = Jf(x) · Jg(u)(du1, · · · ,dum) T (x = g(u)) = Jf(d(u)) · (dg1, · · · ,dgn) T 11
这个等式空为全微分的形式明经性22例 5 5u=log求du及u的偏导数Va?+y?解22du=d logd (logz2 -ln(r? +y))82+92021d(r?+y)2+yz-2r2y21zdrdy+?+y2 +y2z这说明uorau2you_2ar22+y2ay22 +y2,3=284中值公式与Taylor公式果p,qE",令o(t) =(1-t)-p+t·q, teR.我定空α：[0,1]→Rn为R"过从接p,的故参段果A为Rn过的子素，如设Va1,a2EA，从接a1,a2的故参段故包时于A，则空A为凸素。特别螺，凸素都是道路从通的.我定把开的凸素空为凸域例1Rn过开球Br(a）都是凸素定理1（旋分中值定理）果D为Rn过凸域f：D→R可微,则Va,beD,SED,使得f(b) -f(a) = Jf() (b - a)分过=a+0(b-a),E(0,1),位于从接a,b的故参段上证明果[0,1]→D为从接a,b的故参段,则复合函数(t)=fo(t)是可微的一元函数，由一元函数的微分过值们理，日0E（0,1）使得(1) - (0) = (0) ~ (1 - 0),上式即f(b) -f(a) = Jf() (b - a)12
RQ2g SG1 gÆ!. 5 u = log z 2 p x2 + y 2 , N du u u 1E-p. du = d log z 2 x2 + y 2 = d logz 2 − ln(x 2 + y 2 ) = 2 z dz − 1 x2 + y 2 d(x 2 + y 2 ) = −2x x 2 + y 2 dx − 2y x 2 + y 2 dy + 2 z dz. Rq9 ∂u ∂x = − ∂x x2 + y 2 , ∂u ∂y = − 2y x2 + y 2 , ∂u ∂z = 2 z . §4 :9 5 Taylor a p,q ∈ R n , . σ(t) = (1 − t) · p + t · q, t ∈ R. Æ7 σ : [0, 1] → R n R n b% p,q 1Y;. a A R n b1it, Za ∀a1,a2 ∈ A, % a1,a2 1Y;YcB A, N A t. {3, t9j //%1. Æ71t E. 1 R n bM Br(x) 9jt. z 1 (&:9z) a D R n bE, f : D → R , N ∀ a,b ∈ D, ∃ ξ ∈ D, e0 f(b) − f(a) = Jf(ξ) · (b − a), Gb ξ = a + θ(b − a), θ ∈ (0, 1), ξ B% a,b 1Y;]. 8 a σ : [0, 1] → D % a,b 1Y;, NIidp ϕ(t) = f ◦ σ(t) j 1/Gdp. >/Gdp1GbZ7 , ∃ θ ∈ (0, 1) e0 ϕ(1) − ϕ(0) = ϕ 0 (θ) · (1 − 0). ]gv f(b) − f(a) = Jf(ξ) · (b − a). 12