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 >1<` (>G 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 <B G 1 k 6= 0, h 6= 0 G0dp ϕ(y) = f(x0 + h,y) − f(x0,y), ψ(x) = f(x,y0 + k) − f(x,y0), > 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%%<` Σ : D → R n (n > 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} ![ (# c u 0 Σ 1UN5), N>R J(P 1Æb Σ(u 0 ) 1i| J |, J|1U A|, A|b1Gt A(. , m = n − 1 c, Σ P8, #cA| p 1. {3, <B R 3 b 1 () P8 Σ, Σ(u,v) = (x(u,v),y(u,v),z(u,v)) 5