教程高数课件

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

¶z=df ¶z=df zf(exyy2)¶z及xy-

xy-解：令u= ，则z=f(u)=f

xy-

x=

¢exy-y2xy-y2= f中间变量求导.

) )zf(u,v),uf(t),v=y(t)的情形定理如果函数uf(t)及v=y(t)都在点t可导函数zf(u,v)在对应点(u,v)具有连续偏导数则复合函数zf[f(t),y(t)]在对应点t可导, dz=¶zdu+¶zdv. ¶udt ¶vdtt则Duf(tDt-f(tDvy(tDt-y(tDzDz=ADu+BDv+o(r)zf(u,v)在点(u,v)有 Dz=¶z Du+¶z Dv+eDu+eDv 1Dt 2Dt当Dtfi0时,Dufi0,DvfiDufidu

Dvfidv dz=limDz= du+ Dtfi0

如zf(u,vw),uu(tvv(tww(tu变量树图z dz=¶z du+¶z dv+¶z dt dt dt dt导数dz称为全导数(又称链 d如zf(u,vw),uu(tvv(tww(tdz=¶z du+¶z dv+¶z dt dt dt dt

中间变量的个数·y(cosx)sinx,求 令u=cosx,v=sin 则y=dy=¶ydu+¶y =vuv-1(-sinx)+uvlnu(cos=(cosx)1+sinx[lncosx-tan2 zf(u,v),ufxy),v=yxy的情形复合函数为zf[fxyyx具有对x和y的偏导数,且函数zf(u,v)在对应点(u,v)具有连续偏导数,则复合函数zf[fxy),yxy)]在对应点xy)偏导数存在,且可用下列¶z=¶z¶u+¶z¶v ¶v

¶u z=f[f(x,y),y(x,

z ¶z= ¶u+ ¶z= ¶u+ 设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+设ufxy),v=yxywwx都在点xy)处具zf[fxy),yxy),wxy)]在对应点x 的两个偏导数存在,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,例设u=f(x2+t2 ),f具有二阶连续偏导数求¶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自变量xt的复合函数

¶r2-x¶r¶s+x4¶s2+x3¶s设设ufx2t2t),f具有二阶连续偏导数求,.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例设ufx,y)的所有二阶偏导数连续 (1)¶x +¶y (2)

+¶y2 xrcosqyr函数ufx,y)换成极坐标r及q的函数u=f(x,y)=f(rcosq,rsinq)=F(r,qufxy)uF(r,q

r=

+y2,

=arctanx

¶2u 现将¶x

+

+

uF(r,q)对r,q的偏导数来表达

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

设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+=

设函数z全微分dz¶zdu¶z 当ufxy),v=yxy)时,

= 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设ufxxyvgxxy),fg均连续可微

=(

+

y)

设zf(2xygxxy其中f(t)二阶可导设¶2zg(u,v)有连续二阶导数,求¶x¶y.设t2xyux,v¶z=f¢2+

1+ ¶2z

=2ft¢(-

+

0+ x++y( 0+ ft+vv设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)),由题设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=思考题设fx,yz)是k次齐次函数,f(txtytztkfx,yz),l为某一常数,正确的是(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 设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具有二阶连续偏导数¶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,证明z=1¶z+1¶z=z

f(x2-y2

x y y2y引入中间变量,令tx2y2,zty为中间变量,xy为自变量

f(t¶z=-2xyft) ¶z= +2y2f(t f2(t

f(t

f2(