教程高數(shù)課件

zf(u)f[ux,y)]fuuxuyyz= y

?z=df ?z=df zf(exyy2)?z及xy-

xy-解：令u= ，則z=f(u)=f

xy-

x=

￠exy-y2xy-y2= f中間變量求導.

) )zf(u,v),uf(t),v=y(t)的情形定理如果函數(shù)uf(t)及v=y(t)都在點t可導函數(shù)zf(u,v)在對應點(u,v)具有連續(xù)偏導數(shù)則復合函數(shù)zf[f(t),y(t)]在對應點t可導, dz=?zdu+?zdv. ?udt ?vdtt則Duf(tDt-f(tDvy(tDt-y(tDzDz=ADu+BDv+o(r)zf(u,v)在點(u,v)有 Dz=?z Du+?z Dv+eDu+eDv 1Dt 2Dt當Dtfi0時,Dufi0,DvfiDufidu

Dvfidv dz=limDz= du+ Dtfi0

如zf(u,vw),uu(tvv(tww(tu變量樹圖z dz=?z du+?z dv+?z dt dt dt dt導數(shù)dz稱為全導數(shù)(又稱鏈 d如zf(u,vw),uu(tvv(tww(tdz=?z du+?z dv+?z dt dt dt dt

中間變量的個數(shù)·y(cosx)sinx,求 令u=cosx,v=sin 則y=dy=?ydu+?y =vuv-1(-sinx)+uvlnu(cos=(cosx)1+sinx[lncosx-tan2 zf(u,v),ufxy),v=yxy的情形復合函數(shù)為zf[fxyyx具有對x和y的偏導數(shù),且函數(shù)zf(u,v)在對應點(u,v)具有連續(xù)偏導數(shù),則復合函數(shù)zf[fxy),yxy)]在對應點xy)偏導數(shù)存在,且可用下列?z=?z?u+?z?v ?v

?u z=f[f(x,y),y(x,

z ?z= ?u+ ?z= ?u+ 設zeusinvuxy,vxy?z和?z 解?z= ?u+ =eusin y+eucosv=exy[ysin(x+y)+cos(x+?z= ?u+ =eusin x+eucosv=exy[xsin(x+y)+cos(x+設ufxy),v=yxywwx都在點xy)處具zf[fxy),yxy),wxy)]在對應點x 的兩個偏導數(shù)存在,且可用下 ?w ?y

?z +?w z

u2+v2+w,u=x2+y2,u2+v2+ww2xy.解?z(u2(u2+v2+w2

+?z?v+?z?z?v ?z (ux+vx+1122-2=-2++w2zf(ux,y),其中ufx,y的情形?w似 似?v= ?w=

?v= ?w= =

= ?u+?f zf[fx,yxzf[fx,yx,中的u及y例zf(uxyeusin(xy而uxy?z解?z

?f

+

x =eusin(x+ y+eucos(x+?z= =eusin(x+ x+eucos(x+ f(r,例設u=f(x2+t2 ),f具有二階連續(xù)偏導數(shù)求?2u ?2u

解

=f

+?f

=2x?f- x

=f￠

?2

?2 -t

=2?r+2x(?r

+2t?f-t(

-t+?2

2x3

x2

2?2 4t?2 t2?2 2t ? ?r自變量xt的復合函數(shù)

?r2-x?r?s+x4?s2+x3?s設設ufx2t2t),f具有二階連續(xù)偏導數(shù)求,.xrs?u=2x?f-t

?2

2t+?2 1)

1

x -t

?2

2t

?2 1 ?2f+?2f-1?f-2t2?2f-t?2=4xt?r

2 x2 x2 x3?s2例設ufx,y)的所有二階偏導數(shù)連續(xù) (1)?x +?y (2)

+?y2 xrcosqyr函數(shù)ufx,y)換成極坐標r及q的函數(shù)u=f(x,y)=f(rcosq,rsinq)=F(r,qufxy)uF(r,q

r=

+y2,

=arctanx

?2u 現(xiàn)將?x

+

+

uF(r,q)對r,q的偏導數(shù)來表達

u=F(r,q

r=x2

+y2,q=arctanx

?u ?q

?r ?q

1?u得?x

+?y=?r

r?r=r?r=

?u

?u ?q ?2u=?

?x(?rcosq-?q =cosq2u

q]

?u(-sinq)r2

+

-rq

-

r

設ufx,y)的所有

?r=

2sinqcosq

cosq

?r?q+

+sin2q?2u+2sinqcosq(2)?x2(2)?x2+

r

?u?2u ?2u=

?y=?rsinq+?q22sinqcosq?2u2sinq

?r= +cos2q?u+cos2q?2u-2sinqcosq

r

r ?2u+?2u=?2u+1?u+1?2u r r (2)?x2+= ??u r2[r?r(2)?x2+=

設函數(shù)z全微分dz?zdu?z 當ufxy),v=yxy)時,

= dx+dy 例已知e d(e-

2zez0,?z和?z -2z+ez)=e-xyd(-xy)-2dz+ezdz=(ez-2)dz=e-xy(xdy+ye- xe-dz=(ez-2)dx+(ez-2)?z=ye-

xe- ez

-2

?y=ez-2設ufxxyvgxxy),fg均連續(xù)可微

=(

+

y)

設zf(2xygxxy其中f(t)二階可導設?2zg(u,v)有連續(xù)二階導數(shù),求?x?y.設t2xyux,v?z=f￠2+

1+ ?2z

=2ft￠(-

+

0+ x++y( 0+ ft+vv設zf(2xyysinx其中f(u?2z

u2xyvysinf￠f￠?2z

=2[f

f

sinx]+cosxf￠+ycosx[1)+sin=-2uu+(2sinx-ycosx)u+ysinxcos+cos

f(1,1)=1,??x

=2, ??

=3,fx)fx,fxx)),由題設f(1=f(1,f(1,1))f(1,1)df3(

2(x)df x=

x=3(x,+(x,f(x,=3[2+3(2+3)]=

x=思考題設fx,yz)是k次齊次函數(shù),f(txtytztkfx,yz),l為某一常數(shù),正確的是(Cx?f+y?f+z?f=klf(x,y, x?f+y?f+z?f=lkf(x,y, x?f+y?f+z?f=kf(x,y, x?f+y?f+z?f=f(x,y, ff(tx,ty,tz)=tkf(x,y,utx,vtywtz,f(txtytztkfx,yff(u,v,w)=tkf(x,y,

?w=ktk-1f(x,y, tx?f+ty?f+tz?f=tktk-1f(x,y,tkf(xtkf(x,y,z)=kx?f+v?f+z?f=kfu,v,z(C)x?f+(C)x?f+y?f+z?f=kf(x,y,

=kf(u,v, 1992 設zf(exsinyx2y2其中f(uv)?2zuexsiny,vx2?2z

=f

exsinx

f￠

=f￠ecosy+ sin

(fu cosy+f

2+2=e2xsinycosyf

+￠2+2ex(ysin+xcosy)f￠+4xyf￠+excosyf f(uv具有二階連續(xù)偏導數(shù)?2f+?2f

f[

1(x2

y2=x2+y2=x2+y2

1,又g(x, 求?2

y+

?g=?fx+?f(-

=y(?2

?2 ?2 ?2 2?2

?2 2?2

=

?v2+?v?2g=2?2 ?2

2?2

-2

已知f(t)可微,證明z=1?z+1?z=z

f(x2-y2

x y y2y引入中間變量,令tx2y2,zty為中間變量,xy為自變量

f(t?z=-2xyft) ?z= +2y2f(t f2(t

f(t

f2(