案例第八章答案

8.2zx3yz3x2yy3zx33y2 s

u2 2u2vu2v

u2v

v2u lnln

x

lnxy;y2

(4)zsinxycos2xyzycosxy2ycosxysinxyycosxyysin2xyzxcosxyxsin2xy

zlntanyz

2x 2解： ycotysecy

z1zy21xyy1 eyln1xy

xy

xy eyln1xyln1xy 1xyyln1xy .

1xy

1xyyuxu

y1uxz;

ylnx

;

yx

ylnx

uarctanx zx zxyz1 xyzlnxy解：x1xy2z;y1xy2z;z 1xy2 zx4y44x2z4x38xy2z4y3

zxyyzy1zxx y zxsinxyzsinxyxcosxyzxcosxy

zarctany x

x2x1

x2y2

2x1x

x2

xyu y zxz

xzxz1

x 解：xyy ;yy2y ;zy

y uzuzxyylnzuzxyxlnzu

uu

y2x22x

2x2解： yy

y(1)zxy

xy

xx,1,

解zx,1xzxx,1z1,y1y 1yzy1,y

1y2y1y1y2y1y

y11y21yx2（x2

x1,0,

x解zx0x

x,

x1,00.z0,y0,x

0,1zarctanxy1

x0,0,

解zx0arctanx

0,0 1

1.z1,y,

zxcosyycosx1sinxsin

x0,0,

0,zx,0

x1sin

,

0,0 1sin1sinxxcos1sinz0,y

1sin

,

0,0

1sin1sinyycos1siny

fx,y

x2y2,x2y2x2在00處連續(xù)，但fx0x2

x,

fx,y

x,

0f00fxy在00fx,0xfx00設(shè)ulnx3y3z33xyzuuu xy

u

3x2

,u

3y2

,u

3z2, x3y3z3

x3y3z3

x3y3z3

3x2y2z23yzxz 所 x3y3z3 xy設(shè)xcos,y解

xsincos,ysinsinzcosx,y,z ,,

2sinx,y,z

sin sin

,,

cos cossin sin

sin

2

x2,x2x2fx,y0,x2y2x2yx2y

x2

,x2y2解fxxy

x2

3x2y230,x2y2 3,x

fyx,yx2y2 sin3, fxcos,sin0,

cos3, fycos,sin 0,

所以它們在全

設(shè)uarctan2xt證明x22xtu

82xt

42xt 證明

12xt2,

12xt

2 所22 2 1222 2

證明：函數(shù)u

2a e2a u

2ax2u

2a 4a2a 4a2te 14a4at

x

e4a2t2a 2a

x

e

2at2at

1

4a2t

x

2a 4a2a 4a4t所以u

2x2設(shè)uuxyvvxyDuv,uv y 證明：uux,y,vvx,y在區(qū)域D上滿 方程， u 0,v v

v

y

u

u

u

y

x

v

xx2fx,y

,x2y20,x2y2求fxx00,fyy00并證明fxy00不存在解fx00所以fxx000，由對稱性fyy00x2yx2y

x2

,x2y2 x2 2fxx,ydfx,0

x0,x2y2

y

xf0,y0,yx1,y

y0

00不存在設(shè)T

，求證lTg

lg lgl

g

g2l2l

3lg2l11

2

2zexy

x

y

11

11 112 2

xy

xy xy

x

xe

ye

2e

x24

,在點245處的切線x軸的傾角是多y答：傾角的正切即為z在點24處對x的偏導數(shù)

2,42

1，所以傾角是42z2 2求下列函數(shù)的x2y2和xy（1）zx4y44x2z

4x38xy2

2

28y2

2

16xy.

2

12

2

zarctanxz

x2

1y ,z y

2 ,z2

y y

x2y2

1

x2

x2y22

x2y22y2

y2 2

x2y2

x2y22,y2x2y22

zz

2

2

2

解：

lny,y

y

y,xy

lny

,y2xx1 (4)zlnxy2z

2 2

2 2 2 xy2,yxy2,x2xy22,xyxy22 22z

2xy24xy2

2xy2xy2

uarctanx11xyyxy 1xy

1

x

xy 11xy

1x21y2 1x2 12u

2

1x22,xy0,y21y22 設(shè)fx,y,zxy2yz2zx2,求f0,0,1,f1,0,2,f0,1,0及 fx0,1x2f00,1 fx0zzx2fx0z2zx,f10z2z,f102 f0yzyz2 f0,y,zz2,f0,1,zz2,f0,1,z2z,f0,1, fx,0,zzx2,所以fx,0,zx2,fx,0,z0, 2,0,1 3z 3zxlnxy，求x2y及xy2z

2

2

解： lnxy1,x2x,x2y0,xyy,xy2y2 設(shè)usinx2y2求u3u3 u2xcosx2y2,u 2cosx2y24x2sinx2y2

12xsinx2y28x3cosx2y2

12ysinx2y28y3cosx2y2 設(shè)uxxmy

n,m,n為正整數(shù)，求 xmxx解： xm 2yekntsinnx滿

kx2

sinnx,

2

sinnx所以

2ykx2x2y2x2y2r 滿 x2y2x2y2

x rx

r2x2 r2

r2

x

,x2 r2

,y2

z2 2r2r