太原理工微積分與數(shù)學(xué)模型10年修改版第九章理工大高數(shù)

第四節(jié)多元復(fù)合函數(shù)求導(dǎo)法則~

鏈?zhǔn)椒▌t二 全微分形式不變性1.復(fù)合函數(shù)的中間變量為一元函數(shù)的情形定理1如果函數(shù)u

=j

(t)及v

=y

(t)都在點(diǎn)t

可導(dǎo)，函數(shù)z

=f

(u,v)在對(duì)應(yīng)點(diǎn)(u,v)具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)z

=f

[j

(t),y

(t)]在對(duì)應(yīng)點(diǎn)t可導(dǎo)，且其導(dǎo)數(shù)可用下列公式計(jì)算dz

=

?z du

+

?z

dvdt

?u

dt

?v

dt一、鏈?zhǔn)椒▌t?u

?v1

2Dz

=

?z

Du

+

?z

Dv

+

e

Du

+

e

Dv當(dāng)Du

fi

0,

Dv

fi

0

時(shí)，e1

fi

0,

e2

fi

0Dz

=

?z

Du

+

?z

Dv

+

e

Du

+

e

DvDt

?u

Dt

?v

Dt

1

Dt

2

Dt當(dāng)Dt

fi

0時(shí)，Du

fi

0,Dv

fi

0證:設(shè)t

獲得增量DtDu

=

j

(t

+

Dt

)

-j

(t

)，Dv

=y

(t

+

Dt

)

-y

(t

)由于函數(shù)z

=f

(u,v)在點(diǎn)(u,v)有連續(xù)偏導(dǎo)數(shù)dt

?u

dt

?v

dtDt

fi

0

Dtdz

=

lim

Dz

=

?z如上定理的結(jié)論可推廣到中間變量多于兩個(gè)的情況dz

=

?z

du

+

?z

dv

+

?z

dwdt

?u

dt

?v

dt

?w

dtuv

twzDt

dtDu

fiDt

dtdu

+

?z

dvdu

,

Dv

fi

dv全導(dǎo)數(shù)解：dz

=

?z du

+

?z dv

+

?z

dtdt

?u

dt

?v

dt

?t

dt=

vet

-

usin

t

+

cos

t=

e

t

cos

t

-

e

t

sin

t

+

cos

t=

e

t

(cos

t

-

sin

t

)

+

cos

t例1設(shè)z

=uv

+sin

t,求全導(dǎo)數(shù)v

=

cos

tu

=

e

t

,dtdzz

=

f

[f(

x,

y),y

(

x,

y)]2.復(fù)合函數(shù)的中間變量為多元函數(shù)的情形定理2如果u

=j

(x,y)及v=y

(x,y)都在點(diǎn)(x,y)具有對(duì)x

和y

的偏導(dǎo)數(shù)，且函數(shù)z

=f

(u,v)在對(duì)應(yīng)點(diǎn)(u,v)具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)z

=f

[j

(x,y),y

(x,y)]在對(duì)應(yīng)點(diǎn)(x,y)的兩個(gè)偏導(dǎo)數(shù)存在，且可用下列公式計(jì)算uvxzy鏈?zhǔn)椒▌t如圖示?z

=

?z

?u

+

?z

?v?x

?u

?x

?v

?x?z

=

?z

?u

+

?z

?v?y

?u

?y

?v

?yzuvwxy?w

?y+=

+?y

?u

?y

?v

?y?w

?x?z

?w?x

?u

?x

?v

?x?z

?z

?u

?z

?v?z

=

?z

?u

+

?z

?v

+

?z

?w類(lèi)似地再推廣，設(shè)u

=j

(x,y)，v

=y

(x,y)w

=w

(x,y)都在點(diǎn)(x,y)具有對(duì)x

和y的偏導(dǎo)數(shù)，復(fù)合函數(shù)z

=f

[j

(x,y),y

(x,y),w

(x,y)]在對(duì)應(yīng)點(diǎn)(x,y)兩個(gè)偏導(dǎo)數(shù)存在，且可用下列公式計(jì)算解:例2設(shè)z

=eu

sin

v

，u

=xy

，v

=x

+y求?z

，?z?x?y?x

?u

?x

?v

?x=

eu

sin

v y

+

eu

cos

v

1=

e

xy

[

y

sin(

x

+

y)

+

cos(

x

+

y)]?z

=

?z

?u

+

?z

?v?z

=

?z

?u

+

?z

?v?y

?u

?y

?v

?y=

eu

sin

v x

+

eu

cos

v

1=

e

xy

[

x

sin(

x

+

y)

+

cos(

x

+

y)]?f?x?z

=

?f

?u

+?x

?u

?x?f?y?u

?y=

?f

?u

+?z?y即

z

=

f

[f(

x,

y),

x,

y]

令v

=

x

，w

=

y?x?v

=

1

,?x?w

=

0

,?y?v

=

0

,?y?w

=

1把復(fù)合函數(shù)z

=f

[f(x,y),x,y]中的y看作不變而對(duì)x的偏導(dǎo)數(shù)把z

=f

(u,x

,y

)中的u

及y

看作不變而對(duì)x

的偏導(dǎo)數(shù)兩者的區(qū)別區(qū)別類(lèi)似特殊地z

=f

(u,x,y)其中u

=f(x,y)解：

?u

=

?f?x

?x?(

xyz

)

?xdx

?(

2

x

-

y

)?f

?(

xyz

)?x?(

2

x

-

y

)dx

+

?f+例3

設(shè)u

=

f

(

x,2

x

-

y,

xyz)求?x?u=

f1

+

2

f2

+

yzf3解：令u

=x

+y

+z

,v

=

xyz記,?f

(u,

v)f1￠=?u?u?v?2

f

(u,

v)f1￠2￠=f22同理有f2

,

f11

,?u

+

?f?w

=

?f?x?u

?x

?v

?x21?v

=

f

+

yzf例4設(shè)w

=f

(x

+y

+z,xyz)，f

具有二階連續(xù)偏導(dǎo)數(shù)，求和?x

?x?z?w

?2

w=?x?z?2

w?z??z?f?z?f(

f1￠+

yzf2￠)

=

1

+

yf2￠+

yz

2?f1

=

?f1?z

?u?v

=

f?u

+

?f1?z

?v

?z?u

+

?f21211+

xyf?f2

=

?f2?z?2

w?u

?z

?v

?z?v

=

f2221+

xyf=

f11

+

xyf12

+

yf2

+

yz(f21

+

xyf22

)2=

f11

+

y(

x

+

z)

f12

+

xy

zf22

+

yf2?x?z二、全微分形式不變性設(shè)函數(shù)z

=f

(u,v)具有連續(xù)偏導(dǎo)數(shù)，則有全微分dz

=?z

du

+?z

dv

，當(dāng)u

=j

(x,y)?u

?vv

=y

(x,y)時(shí)，有dz

=?z

dx

+?z

dy?x

?y全微分形式不變形的實(shí)質(zhì)無(wú)論z

是自變量(x,y)的函數(shù)或中間變量(u,v)的函數(shù)，它的全微分形式是一樣的。

?u

?x

?v

?x

dz

=

?z

dx

+

?z

dy?x

?y?u

+

?z

?v

dy

?u

?y

?v

?y=

?z

?u

+

?z

?v

dx

+

?z

?u

?x

?y

?y?v

?x=

?z

?u

dx

+

?u

dy

+

?z

?v

dx

+

?v

dy=

?z

du

+

?z

dv?u

?v?z?x

=

e

z

-

2?zye

-

xy

xe-

xy?y

=

ez

-

2例5求

和?x

?y解：

d

(e-

xy

-

2z

+

ez

)