高數(shù)大一下課件73全微分

total小 思考 作第七章多元函數(shù)微分學(xué)1第七章多元函數(shù)微分學(xué)2

的定義先來(lái)介紹全增量設(shè)?元函數(shù)zf(xy)在點(diǎn)P(xy)內(nèi)有定義當(dāng)變量x、y在點(diǎn)(xy)ΔxΔy時(shí),Dz=f(x+Dx,y+Dy)-f(x,f(xy)在點(diǎn)(xy)的全增量3y=f(Dy=f(x+Dx)-f(x)=ADx+則稱函數(shù)y=f(x)在點(diǎn)x可微dy=AΔx=f(4二元函數(shù) 定如果函數(shù)zf(xy)在點(diǎn)(xy)DzfxDx,yDyfx,y)ADx+DzADx+(Dx)2+(Dy)2其中AB僅與x(Dx)2+(Dy)2r

則稱函數(shù)zf(xy)在點(diǎn)(x處可微分,ADxBDy稱為函數(shù)z=f(xy)(xy)

記作dz,dz=ADx+函數(shù)若在某平?區(qū)域D內(nèi)處處可微時(shí)這函數(shù)在D內(nèi)的可微函數(shù).

5ddz=ADx+ADx+ADx+注Dz Dz與dz之差是比r?階?窮?微分系數(shù)A=?B=?6 ( 如果函數(shù)z=f(x,y)在(xy可微分則該函數(shù)在點(diǎn)(xy)的偏導(dǎo)數(shù)?z?x必存在,且函數(shù)z=f(x,y)在點(diǎn)(x,y) 7偏導(dǎo)數(shù)?偏導(dǎo)數(shù)?z?z必存在且函數(shù)zf(xy)在點(diǎn)(xy)如果函數(shù)zf(xy)在點(diǎn)(xy)可微分則該函數(shù)在點(diǎn)(xy)為 證z=f(xy)在點(diǎn)P(xy可微分PxDxyDy?P

Dz=Dz=f(x+Dx,y+Dy)=ADx+BDy+o(r當(dāng)Dy0時(shí)上式仍成立,此時(shí)r=|DxDz=f(x+Dx,y+0)-f(x,y)=+?z=limf(x+Dx,y)-f(x,y)=

Dxfi 可微B可微88 ,多元函數(shù)在某點(diǎn)各個(gè)偏導(dǎo)數(shù)即使都存在,都不能保證函數(shù)在該點(diǎn)連續(xù).答:如果函數(shù)zf(xy)在點(diǎn)(xy可微分,數(shù)在該點(diǎn)連續(xù)事實(shí)上

定義limDyDxfiDz=AD定義limDyDxfi可得limDzlim(ADxBDyo(r))rfi rfi顯然不連續(xù)的函數(shù)?定是不可微的9

存在由定理1知:?元函數(shù)可微?定存在兩個(gè)偏導(dǎo)數(shù) x2+如,fx2+

x2+y2?. x2+y2=在點(diǎn)(0,0)處有,fx(0,0)fy(0,0)(由偏導(dǎo)數(shù)定義可求得 Dz=ADx+BDy+o(r)Dz-[fx(x0,y0)Dx+fy(x0,y0)Dy]?o(rDz-[f

(0,0)Dx+

y(0,0)Dy]

Dx (Dx)2+(Dy)2P(Dx(Dx)2+(Dy)2(Dx)2(Dx)2+(Dy)2

DxDx (Dx)2+(Dx)2 說(shuō)明它不能隨著rfi0?趨于0,當(dāng)rfi0時(shí)Dz-[fx(0,0)Dx+fy(0,0)Dy]?o(r因此,函數(shù)在點(diǎn)(0,0)處不可微存在存在?個(gè)原則區(qū)別現(xiàn)再假定函數(shù)的各個(gè)偏導(dǎo)數(shù)連續(xù)則可證明定理2(微分充分條件)如果函數(shù)z=f(x,y)偏導(dǎo)數(shù)?z?z在xy)連續(xù)則該函數(shù)在點(diǎn)(x 可微分證假定偏導(dǎo)數(shù)在點(diǎn)P(xy)連續(xù)就含有在該點(diǎn)的某?鄰域內(nèi)必存在的意思.(今后常這樣理解Dz=f(x+Dx,y+Dy)-f(x,=[f(x+Dx,y+Dy)-f(x,y+

+[f(x,y+Dy)-f(x,f(x+Dx,y+Dy)-f(x,y+=fx(x+q1Dx,y+=fx(x,y)Dx+1Dx

由由fxxy)在點(diǎn)xy)連續(xù)令fxx+q1DxyDyfxxy其中e10(Dx0,Dyfi同理fx,yDyfx,fyxy)Dye2Dy當(dāng)Dyfi0時(shí),e2fiDz=fx(x,y)Dx+e1Dx+fy(x,y)Dy+=fx(x,y)Dx+fy(x,y)Dy+1Dx+e2Dz-[fx(x,y)Dx+fy(x,y)Dy]=e1Dx+r￡

+

fizf(xy)在點(diǎn)(xy)處可微 定理2的條件(即兩個(gè)偏導(dǎo) 在點(diǎn)(x, 連續(xù))僅是函數(shù)zf(xy)在點(diǎn)(xy)處可微的充條件,并非必條件. 函數(shù)fx,y)

(x2+y2x2

+

x2+y2?在原點(diǎn)(0,0)可微事實(shí)上

x2+y2=(Dx)2 fx

Dxfi

f(0+Dx,0)-f

Dxfi

(Dx)2Dz-[fx(0,0)DxDz-[fx(0,0)Dx+fy(0,0)Dy]=o(rr=(Dx)2+(Dy)2f(x,y)=函 f(x,y)=

x2+y2?x2+x2+y2=fx(0,0)=fy(0,0)=Dz=f(0fx(0,0)=fy(0,0)=rfi

=[(Dx)2+(Dy)2r

(Dx)2+(Dy)2[fx(0,0)Dx+fy(Dx)2+(Dx)2+(Dy)2limrsin10于是,Dz-f(0,0)Dxfrfi r2

DzDz-[fx(0,0)Dx+fy(0,0)Dy]=o(rr=(Dx)2+(Dy)2f(0,0)=x函數(shù)fx,y)(x2+y2) xf(0,0)=x函數(shù)fx,y)(x2+y2) x2+f(0,0)=yx2+y2?x2+y2=f(x,y)在原點(diǎn)(0,0)可微但是偏導(dǎo)數(shù)在原點(diǎn)(0,0)不連續(xù)事實(shí)上,當(dāng)x2y20時(shí)f(x,y)=2x -2 x2+ x2+ x2+特別是當(dāng)yx時(shí),limfx(x,x)=lim(2x

11x21xfi

xfi

2同理可證,fy(x,y)在原點(diǎn)(0,0)也不連續(xù)同理可證,fy(x,y)在原點(diǎn)(0,0)也不連續(xù)函數(shù)在?點(diǎn)可微在這點(diǎn)偏導(dǎo)數(shù)不?定連續(xù)判別f(x,y)在點(diǎn)(x0,y0)是否可微的?法若f(xy)在點(diǎn)(x0y0)不連續(xù)或偏導(dǎo)不存在若f(xy)在點(diǎn)(x0y0)連續(xù)必可微(x-x0)2+(y-y0fxyf(x-x0)2+(y-y0

,

)

?窮??即檢查Dzfxx0y0Dxfyx0y0是否為r的?階?窮?(即極限為0)?若為0,則可微,zf(xy)在點(diǎn)P(xy)處Δz=AΔx+BΔy+o(rΔz=fx(x,y)Δx+fy(x,y)Δy+o(rΔzfxxy)Δxfyxy)Δyfi0(當(dāng)rfir?元函數(shù)的許多微分性質(zhì),這?仍適?.同樣有:(?階) 形式的不變性.★習(xí)慣上

通常把?元函數(shù) 等于它的兩個(gè)偏微之和稱為?元函數(shù)的微分符合疊加原理如三元函數(shù)u=f(x,y,z),則 例 ?z=2x+yexy ?z=xexy,所 dz

dx

求函數(shù)z=ycos(x-2y),當(dāng)x=π,y=dx

π,dy=π時(shí)的 ?z=-ysin(x-2?z=cos(x-2y)+2ysin(x-2π

dx

dy

2π(4-7π).8u的 xu的

.yzxz-

zx x du=

dy+

yyy

yy

y

設(shè)z=f(3x2y), dz= (3zx23xyy2x從2變到2.05,y從3變到試比較Dz與dz的值

x0=

y0=DDx=解Dzfx0Dxy0Dyfx0y0=[(2.05)2+3·2.05·2.96-(2.96)2[22+3·2·3-32=dz=zx(2,3)·0.05+zy(2,3)·(-

y=-三 在近似計(jì)算中的應(yīng)由?元函數(shù) 的定義及關(guān) 存的充分條件可知當(dāng)?元函數(shù)zf(xy)在點(diǎn)P(xy)的兩個(gè)偏導(dǎo)數(shù)fx(x,y),fy(x,y)連續(xù),并且|Dx|,|Dy|都較?時(shí),DDz?dz=fx(x,y)Dx+fy(x,y)Dff(x+Dx,y+Dy)?f(x,y)+fx(x,y)Dx+fy(x,y)DDx=DyDx=Dy= 設(shè)z=f(x,y)=xy,利?函數(shù)f(x,y)=x在點(diǎn)x0y0=(12)處的可微性,(1.04)2.02=f(1.04,2.02)=f(1,2)f(1,2)+=f(1,2)+fx(1,2)Dx+fy(1,當(dāng)兩個(gè)偏導(dǎo)數(shù)fx(當(dāng)兩個(gè)偏導(dǎo)數(shù)fx(xyfy(xy)連續(xù),|Dx|,|Dy|f(x+Dx,y+Dy)?f(x,y)+fx(x,y)Dx+fy(x,y)D 設(shè)函數(shù)fx,y)

x2

(x,

(x,y)=(A)極限不存在(C)可微分

(B)不連續(xù)(Dfx(0,0),fy(0,0)存在設(shè)函數(shù)fx,y)

|xy|,則可得fx(0,0)0,fy(0,0)從?f(xy)在點(diǎn)(0,0)

是零

(非事實(shí)上,設(shè)函數(shù)z=|xy|, |DxDy在點(diǎn)(0,0)處有Dz-[fx(0,0)Dxf|DxDy||DxDyrfi

=Δy=Δy=Dy

0

(Dx)2((Dx)2(Dx)2+(Dx)22=2

|Dx22|Dx 不存在(注意:與?元函數(shù)有很?的區(qū)別可 可導(dǎo)連續(xù)有極偏導(dǎo)連續(xù)可微連續(xù)有極限考研數(shù)學(xué)?,3 f(x,y)的下?