第十二章多元函数的微分81方向导数和偏导数定义1（方向导数）设f：D→R为定义在Rn中开集D内的映射（多元函数).对于pED以及Rn中单位向量u,极限f(p+tu) -t(p)limaf如果存在，则称于在p处沿方向u有方向导数，极限记为-(p),称为f沿u的u方向导数注(1)方向导数就是一元函数g(t)=f(p+tu)在t=0处的导数：特别地of当u=e=(0,…，0,1,0,…,0)（第i个位置为1的单位向量）时，又记(pauaf为(p)，称为的第i个偏导数.按定义,有ariaff(pi,*.,Pi-1,pi +t,pi+1,,Pn) - f(pi,.,Pn)-(p) = limari4taf%仍然可求偏导数，则记=又记为如果”=(2)偏导数ariaa(afaf以及高阶偏称为2阶偏导数。类似地可以定义fiy:=5OrayiOuiar:导数。如果f存在直到k阶的连续偏导数，则称f为Ck函数。我们也使用形如这样的记号：a2faa2fa(af)ofr,(o)OyiOriOyiOri例1求f(a,y)=ry的1阶,2阶偏导数af-a= % -(cd) = , μ= , /= 1, /g = 1, μ= " = 0. 解f(a,y)=例2求f(c,y)=2+y?+ry在(ao,yo)=(1,2)处的偏导数af.afy解(r,y) =(a,y) =十→+yVr2 +y2Vr2+y?V5+2.%(1,2) = )(1,2) =2 V5 +1.Or55Oy1
y }7 {6!x' §1 *u u z1 1 (*u ) a f : D → R 76M R n bt D >1G dp). <B p ∈ D 2u R n b* ( u, s lim t→0 f(p + tu) − t(p) t Za&M, N f M p "+D u ?D-p, sx ∂f ∂u(p), f + u 1 D-p. ; (1) D-pj/Gdp ϕ(t) = f(p+tu) M t = 0 "1-p. {3, , u = ei = (0, · · · , 0, 1, 0, · · · , 0) (4 i Q _ 1 1* () c, Ax ∂f ∂u(p) ∂f ∂xi (p), f 14 i QE-p. 76, ? ∂f ∂xi (p) = lim t→0 f(p1, · · · ,pi−1,pi + t,pi+1, · · · ,pn) − f(p1, · · · ,pn) t . (2) E-p ∂f ∂xi Ax f 0 xi . Za f 0 xi = ∂f ∂xi YVNE-p, Nx f 00 yixi = ∂ ∂yi ∂f ∂xi , 2 E-p. s3276 f 00 xiyi = ∂ ∂xi ∂f ∂yi 2uOE -p. Za f &MY. k 1%%E-p, N f C k dp. Æ7.e= ZR,1xf: ∂ 2f ∂x2 i = ∂ ∂xi ∂f ∂xi , ∂ 2f ∂xi∂yi = ∂ ∂xi ∂f ∂yi , · · · 1 N f(x,y) = xy 1 1 , 2 E-p. f 0 x (x,y) = ∂f ∂x = ∂ ∂x(xy) = y, f 0 y = x, f 00 yx = 1, f 00 xy = 1, f 00 xx = f 00 yy = 0. 2 N f(x,y) = p x2 + y 2 + xy M (x0,y0) = (1, 2) "1E-p. ∂f ∂x(x,y) = x p x2 + y 2 + y, ∂f ∂y (x,y) = y p x2 + y 2 + x ⇒ ∂f ∂x(1, 2) = √ 5 5 + 2, ∂f ∂y (1, 2) = 2 5 √ 5 + 1. 1
例3设(ro,90,z0)ER3,求f(r,9,2) = [(μ - ro)2 + (y - yo)2 + (z - 20)]-±的偏导数解记r = [(r - zo) + (y - yo)2 + (z - 20)j 则oraf--1.91 aTO-Cor2Orr3r同理,af--yoaf_Z-20Jy73zr3和一元函数不同的是，偏导数的存在不能保证多元函数的连续性，这是因为偏导数只反映函数沿特定方向的性质例4设当r·y=of(r,y)r.o则of.f(r, 0) -f(0, 0)(r,0)= lim =0dr201(0,0) = jmf(0. ) = f(0, 0) = 0.dy-0y显然，f在（0,0）处不连续定理1（复合求导）设f如定义1,z°D.假设f(1≤i≤n)在r°附近连续，则(1）f在°处连续(2)如果ai=;(t)在to处可导，ro=(ai(to),,ro(to),则f(r(t))在t=to处可导，且dt=2%(0) -4(o).dtlt=toOri-12
3 a (x0,y0,z0) ∈ R 3 , N f(x,y,z) = (x − x0) 2 + (y − y0) 2 + (z − z0) 2 − 1 2 1E-p. x r = (x − x0) 2 + (y − y0) 2 + (z − z0) 2 1 2 N ∂f ∂x = − 1 r 2 · ∂r ∂x = − 1 r 2 x − x0 r = − x − x0 r 3 . , ∂f ∂y = − y − y0 r 3 , ∂f ∂z = − z − z0 r 3 . g/Gdp1j, E-p1&M?V>Gdp1%%!, Rj7 E-p\B<dp+{7D1!a. 4 a f(x,y) = 0, , x · y = 0, 1, , x · y 6= 0, N ∂f ∂x(x, 0) = lim x→0 f(x, 0) − f(0, 0) x = 0, ∂f ∂y (0, 0) = lim y→0 f(0,y) − f(0, 0) y = 0. V, f M (0, 0) "%%. z 1 (u) a f Z76 1, x 0 ∈ D. za f 0 xi (1 ≤ i ≤ n) M x 0 K %%, N (1) f M x 0 "%%; (2) Za xi = xi(t) M t0 "-, x 0 = (x1(t0), · · · ,x0(t0)), N f(x(t)) M t = t0 "-, K dt dt t=t0 = Xn i=1 ∂f ∂xi (x(t0)) · dxi dt (t0). 2
证明(1)利用微分中值定理，有f()-f(ro) =E[f(rl,.,r-,a,...,an)-f(rl,...r-,a,ai+,..,an)]i=1nfr,(rl,..,rl-i,si,ai+1,...,rn).(ci-rg)i-1当→2o时,→,由f在ro处连续知lim(f()-f(r0))=0.(2)由(1)的证明,有df f(r(t)) -f(r(to)limdtt=tot-tot-to(,, + n)(t)lim>t-tot7ft,(r0) -r;(to)i=1续论在定理1的条件下，如果u=ui=ui,u2...,un),则i=1%(p) = dl-0/0+ ) =≥%(a) 4:OuDr定理2 (向导次序的可交为性)设f：D→R为二元函数，（co,o)ED.如果fu和fu在（royo）处连续则fry(o,yo)=fyr(o,y)证明对于充分小的k≠0,h≠0分例考虑函数p(y)=f(ro+h,y)-f(ro,y),d(r) = f(r,yo +k) - f(r, yo),由Lagrange中值定理，有(yo+k)-(yo)=(yo+0ik)k(1i/≤1)=[f(ro+h,yo+o)-f(ro,1+ok)] k=fiu(ro+02h,y0+0ik).kh(102l≤1)3
8 (1) "=GbZ7 , ? f(x) − f(x 0 ) = Xn i=1 f(x 0 1 , · · · ,x0 i−1 ,xi , · · · ,xn) − f(x 0 1 , · · · ,x0 i−1 ,x0 i ,xi+1, · · · ,xn) = Xn i=1 f 0 xi (x 0 1 , · · · ,x0 i−1 ,ξi ,xi+1, · · · ,xn) · (xi − x 0 i ) , x → x 0 c, ξi → x 0 i , > f 0 xi M x 0 "%%W lim x→x0 (f(x) − f(x 0 )) = 0. (2) > (1) 1V9, ? df dt t=t0 = lim t→t0 f(x(t)) − f(x(t0)) t − t0 = lim t→t0 Xn i=1 f 0 xi (x 0 1 , · · · ,x0 i−1 ,ξi ,xi+1, · · · ,xn) · x(t) − xi(t0) t − t0 = Xn i=1 f 0 xi (x 0 ) · x 0 i (t0). % M7 1 1~, Za u = Xn i=1 ui · ei = (u1,u2, · · · ,un), N ∂f ∂u(p) = d dt|t=0f(p + tu) = Xn i=1 ∂f ∂xi (p) · ui . z 2 (us.w -) a f : D → R @Gdp, (x0,y0) ∈ D. Z a f 00 xy g f 00 yx M (x0,y0) "%%, N f 00 xy(x0,y0) = f 00 yx(x0,y0). 8 Lagrange bZ7 , ? ϕ(y0 + k) − ϕ(y0) = ϕ 0 y (y0 + θ1k)k (|θ1| ≤ 1) = f 0 y (x0 + h,y0 + θk) − f 0 y (x0,y1 + θ1k) k = f 00 xy(x0 + θ2 · h,y0 + θ1k) · kh (|θ2| ≤ 1). 3
同理,d(ro +h)-b(ro)=fur(ro+03h,yo +oyk)hk易见，p(yo+k)-p(yo)=(ao+h)-(ro),故fry(ro+0zh,yo+0ih)=fur(ro+0gh,yo+4k)令k,h→0,由fuy，fu在（ro,o）处连续即得欲证等式推论多元函数的各阶偏导数如果连续，则其值与求导次序无关例 52-9?(r,y) + (0, 0),t22+92f(r,y) =0,(r,y) = (0, 0).则fy(0,0)=1，fu(0,0)=-1，这说明定理2中连续性假设是不可缺少的82切线和切面设α:[α,β]→Rn为Rn中一条连续曲线,记a(t)=(ri(t),.…:,an(t).如果z;(t)(1≤i≤n)在t=to处均可导，则称。在to处可导，记4(0)dag(to) = =(ri(to),...,r(to))dtlt=to称(to)为α在to处的切向量当(to)0时，称((to)+(to)uluER)为在to处的切线，其方程可写为P-o(to)=u·o(to)或2o920i(to)(to)(to)经过α(to）且与切线正交的超平面称为法面，其方程为(q-o(to)) +o'(to) = 04
, ψ(x0 + h) − ψ(x0) = f 00 yx(x0 + θ3h,y0 + θ4k)hk. 3~, ϕ(y0 + k) − ϕ(y0) = ψ(x0 + h) − ψ(x0), Y f 00 xy(x0 + θ2h,y0 + θ1h) = f 00 yx(x0 + θ3h,y0 + θ4k). . k,h → 0, > f 00 xy, f 00 yx M (x0,y0) "%%v0FV2g. % >Gdp1RE-pZa%%, NGZDN-$$[. 5 f(x,y) = xy x 2 − y 2 x2 + y 2 , (x,y) 6= (0, 0), 0, (x,y) = (0, 0). N f 00 xy(0, 0) = 1, f 00 yx(0, 0) = −1, Rq97 2 b%%!zajT^1. §2 ) a σ : [α,β] → R n R n b/~%%P, x σ(t) = (x1(t), · · · ,xn(t)). Za xi(t)(1 ≤ i ≤ n) M t = t0 "-, N σ M t0 "-, x σ 0 (t0) = dσ dt (t0) = dσ dt t=t0 = (x 0 1 (t0), · · · ,x0 n (t0)) σ 0 (t0) σ M t0 "1J(. , σ 0 (t0) 6= 0 c, {σ(t0) + σ 0 (t0)u|u ∈ R} σ M t0 "1J, GD P − σ(t0) = u · σ 0 (t0) q x − x0 x 0 1 (t0) = y − y0 x 0 2 (t0) = z − z0 x 0 3 (t0) . Æb σ(t0) KDJU1F8 A8, GD (q − σ(t0)) · σ 0 (t0) = 0. 4
例1设f为一元可微函数，令o(t) = (t,f(t))则α'(to)=(1,f(to)),α在to处切线方程为- to =y-f(to)1f'(to)即y= f(to) +f'(to) (r - to).例2求螺旋线o(t)=(acost,asint,t),tER的切线和法面方程解在t=to处d(to)= (-asinto,a costo,1)故切线方程为r-acosto_y-asintoz-to1-asintoacosto法面方程为-(α-acosto)asinto+(y-asinto)acosto+(z-to).1= 0.设D为Rm中开集，我们称连续映射：D→Rn（n>m）为Rn中的一个参数曲面.设uo=(ul,*,um）D则u-E(ul, ..,ug-i,u, ui+1,.-.,um)为Rm中曲线，称为上的ui曲线如果ui曲线在u处可导，则记(u) =Suiu(uo)，它是该曲线在u°处的切向量如果[u(u)1≤i≤m)线性无关（此时称u°为的正则点)，则称由这些切向量张成的、经过(u)的子空间为切空间，切空间的正交补称为法空间，法空间中的元素称为法向量当m=n-1时，称为超曲面，此时法空间维数为1.特别地，对于R3中的（超）曲面，E(u,v) = (r(u,v),y(u,v),z(u,v)5
1 a f /Gdp, . σ(t) = (t,f(t)) N σ 0 (t0) = (1,f0 (t0)), σ M t0 "JD x − t0 1 = y − f(t0) f 0(t0) v y = f(t0) + f 0 (t0) · (x − t0). 2 N3& σ(t) = (a cost, a sin t, t), t ∈ R 1JgA8D . M t = t0 ", σ 0 (t0) = (−a sin t0,a cost0, 1) YJD x − a cost0 −a sin t0 = y − a sin t0 a cost0 = z − t0 1 , A8D −(x − a cost0)a sin t0 + (y − a sin t0)a cost0 + (z − t0) · 1 = 0. a D R m bt, Æ7%% m) R n b1/Q pP8. a u 0 = (u 0 1 , · · · ,u0 m) ∈ D, N u 7→ Σ(u 0 1 , · · · ,u0 i−1 ,u,u0 i+1, · · · ,u0 m) R m bP, Σ ]1 ui P. Za ui PM u 0 i "-, Nx ∂Σ ∂ui (u 0 ) = Σ 0 ui (u 0 ), yjLPM u 0 "1J(. Za {Σ 0 ui (u 0 )|1 ≤ i ≤ m} 
其切称量为(uo,vo)= (r(uo,vo),yu(uo,vo),zu(uo,vo)E%(uo, vo) = (r(uo,vo),y(uo,vo),z%(uo, vo)如果(uo,vo),%(uo,vo）线性无关，则n =E(uo,vo) xE%(uo,vo)=(yuz-zu-yu, zu-au-ru-zu, auy-yu-a)+0亢为法称量，推而的切平面方程为(P-E(uo, vo) ·n= 0或改程为-r(uo, vo) y-y(uo, vo) z-z(uo, vo)= 0.zt(uo, vo)ru(uo,vo)yu(uo, vo)z(u0, vo)c(uo, vo)z(uo,vo)例3求球面s2=[(r,y,z)R32+y?+22=1)的切面解球面可程成参数曲面=sincos,y=sinsinp,z=coso0,02,其法称量为n=(coscos,cossinsin)(sinsin,sincos)=sin()故在（co,30,20）处切平面方程为(r - ro) : ro + (y - yo) - yo + (z- zo) . zo = 0.83映射的微分我们回忆一下，对于一元函数而言，可微是关该函数可等和线性函数一阶逼近对于多元函数，我们也可等通过线性逼近来定义可微性6
GJ( Σ 0 u (u0,v0) = (x 0 u (u0,v0),y0 u (u0,v0),z0 u (u0,v0) Σ 0 v (u0,v0) = (x 0 v (u0,v0),y0 v (u0,v0),z0 v (u0,v0)) Za Σ 0 u (u0,v0), Σ 0 v (u0,v0) 
设DCRn为开集我们把映射f：D→Rn称为由元向量值函数，写小元量形式为f(1,*..,an)=(fi(a1,.,Tn),..,fm(r1,..,an))为方便起见，以下把欧氏空间中的向量以列向量来表示定的1（微分）设f如上,zo=(ri,,ro)TeD.如果存在m×n阶的矩阵A=(ai)mxn，使得对于o附近的点a，有If()-[f() +A-(-)]ll=o(l-l), →o则称f在°处可微线性映射df(a0): Rn-→RmUHA·V称为于在°处的微元命处1（可微→可导）如果f：D→Rm在z°处可微，则其元量fi(1≤i≤n)在a°处存在方向导数，并且fi(a0)A=ori证明续微元的定义可以看出，如果f在ao处可微，则f在o处连续下面以m=1为例说明方向导数的存在性为此，取单位向量u，由定义，我们有f(o+tu)-f(r)=A(ro+tu-ro)+o(lro+tu-2o)= t.Au+o(lIt)f(a)=A·u,即方向导数存在特下地,这说明uf(r0)=Aei(af. (a),..afA=→OriOrOrr例1设r?y(r,y)(0,0),2 +y2f(a,y) :0,(r, y) = (0, 0).7
a D ⊂ R n t, Æ7G(Zdp, G ( g f(x1, · · · ,xn) = (f1(x1, · · · ,xn), · · · ,fm(x1, · · · ,xn)) D H~, 2Ck|b1(2*( h. z1 1 (&) a f Z], x 0 = (x 0 1 , · · · ,x0 n ) T ∈ D. Za&M m × n 1 S A = (aij )m×n, e076, Æ7? f(x 0 + tu) − f(x 0 ) = A(x 0 + tu − x 0 ) + o(kx 0 + tu − x 0k) = t · Au + o(|t|) Rq9 ∂f ∂u(x 0 ) = A · u, vD-p&M. {3, ∂f ∂xi (x 0 ) = A · ei ⇒ A = ∂f ∂x1 (x 0 ), · · · , ∂f ∂xn (x 0 ) . 1 a f(x,y) = x 2y x2 + y 2 , (x,y) 6= (0, 0), 0, (x,y) = (0, 0). 7
们为f(,)≤，故球(0,0)处推连，且(0,0)=(0,0)=0.如果u=（u1u2）为言位向量，求t3uiu2uiu2 (0,0) =mf(tui,tu2)limQuX0=0 (u2+u)t3i+u然而f球（0,0）处不切微（why?)如果fi（1≤i≤m）的偏导回参存球，求记Jf=（），称为f的1Jacobian.Jf球每一忆的如构成一个映射Jf：D→Rm-n，这里经定把m×n阶例阵视为Rmn中的忆定理1（无微的充分条件）如果Jf球D中存球且它作为映射球z°处推连，求于球处切微证明故以m=1为例来证明由条件，f球zo处推连，i=1,2,.…，n根论微元中如定理，有[f(ri,...,a-,,+1,*,n) -f(l,...,al,r+,,an)]f(r)-f(ro) =i=1Nfr,(rl,,rg-1,r+o.(i-a),ai+1,...,an)(a-r)i=15nEf(0) ( -29) +Ea (ai -2l)i=1i=1其中ai= f",(ri,.,a-1,a+0(ai-),a1,...,an)-fa,(rl,...,an)→0, (i→a)从而f(a) -f(a0)+E(20) (ai-29)0?-1o(/-)即球处切微如果经定把m×n的例阵视为Rmn中的忆，求例阵令切定义是然的范回即，如果A=（aii）mxn，求其范回定义为I/All = 1<<8
7 |f(x,y)| ≤ 1 2 |x|, Y f M (0, 0) "%%, K f 0 x (0, 0) = f 0 y (0, 0) = 0. Za u = (u1,u2) * (, N ∂f ∂u(0, 0) = lim t→0 f(tu1,tu2) t = lim t→0 t 3u 2 1u2 (u 2 1 + u 2 2 )t 3 = u 2 1u2 u 2 1 + u 2 2 . V? f M (0,0) " (why?). Za fi (1 ≤ i ≤ m) 1E-p&M, Nx Jf = ∂fi ∂xj m×n , f 1 Jacobian. Jf M6/51ZW /Q~, f 0 xi M x 0 "%%, i = 1, 2, · · · ,n. TGbZ7 , ? f(x) − f(x 0 ) = Xn i=1 f(x 0 1 , · · · ,x0 i−1 ,xi ,xi+1, · · · ,xn) − f(x 0 1 , · · · ,x0 i ,xi+1, · · · ,xn) = Xn i=1 f 0 xi (x 0 1 , · · · ,x0 i−1 ,x0 i + θ · (xi − x 0 i ),xi+1, · · · ,xn) · (xi − x 0 i ) = Xn i=1 f 0 xi (x 0 ) · (xi − x 0 i ) +Xn i=1 αi · (xi − x 0 i ) Gb αi = f 0 xi (x 0 1 , · · · ,x0 i−1 ,x0 i +θ(xi−x 0 i ),xi+1, · · · ,xn)−f 0 xi (x 0 1 , · · · ,x0 n ) → 0, (xi → x 0 i ) %? f(x) − " f(x 0 ) +Xn i=1 f 0 xi (x 0 ) · (xi − x 0 i ) # ≤ Xn i=1 α 2 i !1 2 · kxi − x 0 i k = o(kx − x 0 k) v f M x 0 ". ZaÆ7 m × n 1Sl R mn b15, NS.76jV1Cp. v, Za A = (aij )m×n, NGCp76 kAk = X 1≤i≤m 1≤j≤n a 2 ij 1 2 . 8
由Schwarz不等式，有VuER".IIA - ~ll ≤ IIAl - IIol, 定理2(复合空导)设△为R过开集，D为Rm过开集g：△→D及f：D→Rn为对射.如果g在u°E△处可微，f在r°=g(u）处可微。则复合对射h=fog：△→Rn在uo处可微，且Jh(uo)= Jf(r)·Jg(uo)面明因为9在uo处可微，故(1)g(u) - g(u) = Jg(uo) (u -wo) + Rg(u, uo)元过R(u,o)=o(u-ul)：、理，因为f在o=g(uo）处可微故(2) f() -f(r0) = Jf(0) ( -20)+ R(,20)元过R()=(l-)由 (1)知, 当 u→ uo)时, g(u)→ g(u)= r0. 以= g(u) 代入(2),得f og(u) -f og(u) = Jf(ro)(g(u) -g(u))+Rf(g(u),g(uo))(3)= Jf(r)Jg(u).(u-uo)+Rfog(u,uo)元过Rfog(u,uo)=Jf(ro)·Rg(u,u)+Rf(g(u),g(u)从而有如别估计IRfog(u,uo)ll≤IJf(r)Rg(u,u)l + /Ry(g(u),g(uo)≤ /Jf(r)l IRg(u, u)I+o(llg(u)-g(u))= o(lu-ul) + o(O(llu-ul)= o(u-ul)从而由(3)及微分的定义知fog在uo处可微且J(og)(u)=Jf(ro)·Jg(u)如果把f，9分例表示成分量形式yi=fi(ri,*,an),i=l,**,m,Tj = gi(u1,,u),j=l,.,n9
> Schwarz 2g, ? kA · vk ≤ kAk · kvk, ∀v ∈ R n . z 2(u) a ∆ R l bt, D R m bt, g : ∆ → D u f : D → R n (1) W, , u → u 0 ) c, g(u) → g(u 0 ) = x 0 . 2 x = g(u) )[ (2), 0 f ◦ g(u) − f ◦ g(u 0 ) = Jf(x 0 )(g(u) − g(u 0 )) + Rf (g(u),g(u 0 )) = Jf(x 0 ) · Jg(u 0 ) · (u − u 0 ) + Rf◦g(u,u0 ) (3) Gb Rf◦g(u,u0 ) = Jf(x 0 ) · Rg(u,u0 ) + Rf (g(u),g(u 0 )) %??ZXw kRf◦g(u,u0 )k ≤ kJf(x 0 ) · Rg(u,u0 )k + kRf (g(u),g(u 0 )k ≤ kJf(x 0 )k · kRg(u,u0 )k + o(kg(u) − g(u 0 )k) = o(ku − u 0 k) + o(O(ku − u 0 k)) = o(ku − u 0 k). %?> (3) uG176W f ◦ g M u 0 ", K J(◦g)(u 0 ) = Jf(x 0 ) · Jg(u 0 ). Za f, g Gh G( g yi = fi(x1, · · · ,xn), i = 1, · · · ,m, xj = gj (u1, · · · ,ul), j = 1, · · · ,n. 9
则J(fog)(uo）=Jf(r0）Ja(uo）可改写为an(a0)dy1yn(u0)1(r0)or1drr0(uo)aundutdriarmduiJu0yn(u0)Qyn(uo)0yn(r0)ayn(0rm(u0)arm(u0(0OrmdnduidrtduinxlX即oyi(u)u(g(a)ars(u)N-ujOuj=oxs这也就是所谓的链规则下2设f(,y)可微,()可微求u=f(,p()关于的导数解由链规则=f'(c,p(r)):t+f(c,p(r))-p(r)=fi(,(r))+f(,(r)) -p'(r).下3设u=f(c,y)可微，r=rcoso,y=rsino,证明(au)(8u)+1（ou）Ou=Xorarau.2证不由链规则，duauauouarouQy=cos o +- sind+ararararardydyduduardududuaysin o +cOs00ar00ardyay这说明duauauOuOucOsesingsing.arar00dyrdyz下 4 设z= f(u,u,w),u=p(u,s), s=(u,w),求auw10
N J(f ◦ g)(u 0 ) = Jf(x 0 ) · Jg(u 0 ) M ∂y1 ∂u1 (u 0 ) · · · ∂y1 ∂ul (u 0 ) · · · ∂yn ∂u1 (u 0 ) · · · ∂yn ∂ul (u 0 ) n×l = ∂y1 ∂x1 (x 0 ) · · · ∂y1 ∂xm (x 0 ) · · · ∂yn ∂x1 (x 0 ) · · · ∂yn ∂xm (x 0 ) n×m · ∂x1 ∂u1 (u 0 ) · · · ∂x1 ∂ul (u 0 ) · · · ∂xm ∂u1 (u 0 ) · · · ∂xm ∂ul (u 0 ) m×l v ∂yi ∂uj (u 0 ) = Xn s=1 ∂yi ∂xs (g(u 0 )) · ∂xs ∂uj (u 0 ). R.jx 1&_N. 2 a f(x,y) , ϕ(x) , N u = f(x,ϕ(x)) [B x 1-p. >&_N u 0 x = f 0 x (x,ϕ(x)) · x 0 x + f 0 y (x,ϕ(x)) · ϕ 0 (x) = f 0 x (x,ϕ(x)) + f 0 y (x,ϕ(x)) · ϕ 0 (x). 3 a u = f(x,y) , x = r cos θ,y = r sin θ, V9 ∂u ∂x2 + ∂u ∂y 2 = ∂u ∂r 2 + 1 r 2 ∂u ∂θ 2 . 8 >&_N, ∂u ∂r = ∂u ∂x · ∂x ∂r + ∂u ∂y · ∂y ∂r = ∂u ∂x cos θ + ∂u ∂y · sin θ ∂u ∂θ = ∂u ∂x · ∂x ∂θ + ∂u ∂y · ∂y ∂θ = −r · ∂u ∂x · sin θ + r ∂u ∂y cos θ Rq9 ∂u ∂r 2 + 1 r 2 ∂u ∂θ 2 = ∂u ∂x cos θ + ∂u ∂y sin θ 2 + − ∂u ∂x sin θ + ∂u ∂y cos θ 2 = ∂u ∂x2 + ∂u ∂y 2 . 4 a z = f(u,v,w),v = ϕ(u,s), s = ψ(u,w), N ∂z ∂u, ∂z ∂w. 10