復(fù)合函數(shù)的偏導(dǎo)數(shù)課件

1、復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件一、鏈?zhǔn)椒▌t 定理定理如果函數(shù)如果函數(shù))(tuf f= =及及)(tvy y= =t可導(dǎo)，函數(shù)可導(dǎo)，函數(shù)),(vufz = =在對(duì)應(yīng)點(diǎn)在對(duì)應(yīng)點(diǎn)),(vu具有連續(xù)偏具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)導(dǎo)數(shù)，則復(fù)合函數(shù))(),(ttfzy yf f= =在對(duì)應(yīng)點(diǎn)在對(duì)應(yīng)點(diǎn)t可可導(dǎo)，且其導(dǎo)數(shù)可用下列公式計(jì)算：導(dǎo)，且其導(dǎo)數(shù)可用下列公式計(jì)算： dzfdufddtu dtdt =uvtz復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件I: 定理推廣到定理推廣到中間變量多于兩個(gè)中間變量多于兩個(gè)的情況的情況.如如dtdwwzdtdvvzdtduuzdtdz = =uvwtz以上公式中的導(dǎo)數(shù)以上公式中的導(dǎo)數(shù) 稱為稱為d

2、tdz ( ), ( ), ( )zftttfy=)(tuf f= =)(tvy y= =( )wt=復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件定理推廣到定理推廣到中間變量是多元函數(shù)中間變量是多元函數(shù)的情況：的情況：定理定理1(p290).,(),(yxyxfzy yf f= = 如果如果),(yxuf f= =及及),(yxvy y= =都在點(diǎn)都在點(diǎn)),(yx具有對(duì)具有對(duì)x和和y的偏導(dǎo)數(shù)，且函數(shù)的偏導(dǎo)數(shù)，且函數(shù)),(vufz = =在對(duì)應(yīng)在對(duì)應(yīng)點(diǎn)點(diǎn)),(vu具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)),(),(yxyxfzy yf f= =在對(duì)應(yīng)點(diǎn)在對(duì)應(yīng)點(diǎn)),(yx的兩個(gè)偏的兩個(gè)偏導(dǎo)數(shù)存在，且可用

3、下列公式計(jì)算導(dǎo)數(shù)存在，且可用下列公式計(jì)算 ;zzuzvzzuzvxuxvxyuyvy = = = = II:復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件uvxzy鏈?zhǔn)椒▌t如圖示鏈?zhǔn)椒▌t如圖示= = xz uzxu vz,xv = = yz uzyu vz.yv 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件證明證明 證第一個(gè)公式證第一個(gè)公式: ,uxx yx y = = ,vxx yx yy yy y = = 由于由于 Z = f ( u, v ) 在點(diǎn)在點(diǎn) ( u, v ) 可微，有可微，有 =vvfuufZ xxvvfxuufxZ=22 ()():uvx 其其中中，上上式式兩兩邊邊同同除除以以00,00 xuv （ ）令令，則

4、則，從從而而,xxuv 給給以以改改變變量量相相應(yīng)應(yīng)得得到到和和 的的改改變變量量復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件00( )( )lim |lim | |xxxx = = 由由于于2222000( )lim| lim ()( )0 lim ( )( )0 xxxuvuvxxxx = = = = = = 0 limxzzfufvxxuxvx = = = 因因 此此= = yz uzyu vz.yv 同理可證：同理可證：復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件 、III: 再推廣，設(shè)再推廣，設(shè)),(yxuf f= =),(yxvy y= =),(yxww = =都在點(diǎn)都在點(diǎn)),(yx具有對(duì)具有對(duì)x和和y的偏導(dǎo)數(shù)，復(fù)合

5、函數(shù)的偏導(dǎo)數(shù)，復(fù)合函數(shù)),(),(),(yxwyxyxfzy yf f= =在對(duì)應(yīng)點(diǎn)在對(duì)應(yīng)點(diǎn)),(yx的兩個(gè)的兩個(gè)偏導(dǎo)數(shù)存在，且可用下列公式計(jì)算偏導(dǎo)數(shù)存在，且可用下列公式計(jì)算 , zwvuyx;zzuzvzwxuxvxwx = = zzuzvzwyuyvywy= 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件IV:特殊地特殊地),(yxufz = =),(yxuf f= =即即,),(yxyxfzf f= =,xfxuufxz = = .yfyuufyz = = 令令,xv = =, yw = =其中其中, 1= = xv, 0= = xw, 0= = yv. 1= = yw把把),(yxufz = =中中的的u

6、及及y看看作作不不變變而而對(duì)對(duì)x求求偏偏導(dǎo)導(dǎo)數(shù)數(shù) 兩者的區(qū)別兩者的區(qū)別區(qū)別類似區(qū)別類似復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件例例 1 1 設(shè)設(shè)vezusin= =，而，而xyu = =，yxv = =， 求求 xz 和和yz .解解= = xz uzxu vzxv 1cossin = =veyveuu),cossin(vvyeu = = = yz uzyu vzyv 1cossin = =vexveuu).cossin(vvxeu = =復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件例例 2 2 設(shè)設(shè)tuvzsin = =，而，而teu = =，tvcos= =， 求全導(dǎo)數(shù)求全導(dǎo)數(shù)dtdz. 解解dzz duz dvz dw

7、dtu dtv dtw dt=ttuvetcossin = =ttetettcossincos = =.cos)sin(costttet = =sinwt=令 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件(, ),ufx y z= =例例3 3 設(shè)設(shè)解解：u),ln(),(22yxzxy = = = 而而, 可可微微 fdxdu求求xyzxyx= =dxduxfyf dxdy zf xz zf yz dxdy xf= = yfzf 222yxx zf 222yxy xf= = yfzf 22)(2yxyx 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件解解令令, zyxu = =;xyzv = =記記1,ffu=212,ffu v

8、 = 同理有同理有,2f ,11f .22f = = xwxvvfxuuf ;21fyzf = =復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件= = zxw2)(21fyzfz ;221zfyzf yzf = = = zf1zvvfzuuf 11;1211fxyf = = = zf2zvvfzuuf 22;2221fxyf = =于是于是= = zxw21211fxyf 2f y )(2221fxyfyz .)(22221211f yf zxyfzxyf = =(,)wf xyz xyz= 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件 設(shè)設(shè)函函數(shù)數(shù)),(vufz = =具具有有連連續(xù)續(xù)偏偏導(dǎo)導(dǎo)數(shù)數(shù)，則則有有全全微微分分dvvzd

9、uuzdz = =;當(dāng)當(dāng)),(yxuf f= =、),(yxvy y= =時(shí)時(shí)，有有dyyzdxxzdz = =.全微分形式不變形的實(shí)質(zhì)全微分形式不變形的實(shí)質(zhì)： 無論無論 是自變量是自變量 的函數(shù)或中間變量的函數(shù)或中間變量 的函數(shù)，它的全微分形式是一樣的的函數(shù)，它的全微分形式是一樣的.zvu、vu、二、全微分形式不變性復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件dxxvvzxuuz = =dyyzdxxzdz = =dyyvvzyuuz = =dyyudxxuuz dyyvdxxvvzduuz = =.dvvz ( , ),( , ),( , )zf u v ux y vx yfy= 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件

10、解解1, 1= = xu,2cos21yzzeyyu = = ,yzyezu= = 所求全微分所求全微分.)2cos21(dzyedyzeydxduyzyz = =2解解 ：直直接接求求微微分分(sin)()2yzydudxdd e= 1cos22yzyzydxdyzedyyedz 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件解解例例6 6()exyzxy= = 求求函函數(shù)數(shù)的的偏偏導(dǎo)導(dǎo)數(shù)數(shù)和和全全微微分分e)(ddxyyxz = =)(dede)(yxyxxyxy = =)d(de)dd(e)(yxyxxyyxxyxy = =,d)1(ed)1(e22yxyxxyxyxyxy = =所以所以, )1(e2 =

11、 = yxyxzxy.)1(e2 = = xyxyzxy復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件解解例例7 72ln(2 ) zxxy= = 求求函函數(shù)數(shù)的的偏偏導(dǎo)導(dǎo)數(shù)數(shù)和和全全微微分分所以所以)2ln(dd2yxxz = =)2ln(dd)2ln(22yxxxyx = =yxyxxxyx2)2(dd)2ln(222 = =,d22d22)2ln(2222yyxxxyxxyx = =,22)2ln(222yxxyxxz = = .222yxxyz = = 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件例例8 8= =z設(shè)設(shè)、vucossin、xyu = =,xyv = =xz 求求yz 及及解解= =dzvduucoscosv

12、dvusinsin vucoscos= =ydx()xdy vusinsin dxxy2( )1dyx vuycoscos(= =)sinsin2vuxy dxvuxcoscos( )sinsin1vux dyxz vuycoscos= =vuxysinsin2 yz vuxcoscos= =vuxsinsin1 復(fù)合函數(shù)的偏導(dǎo)數(shù)PPT課件三、 高階微分 2()zzzzd zd dzdxdy dxdxdy dyxxyyxy=22222222zzzzdxdydxdxdydyxy xx yy = = 2()dxdyfxy=dyyzdxxzdz = =( , ),( , ),( , )zf u v ux