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 = =在對應(yīng)點在對應(yīng)點),(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= =在對應(yīng)點在對應(yīng)點t可可導(dǎo)，且其導(dǎo)數(shù)可用下列公式計算：導(dǎo)，且其導(dǎo)數(shù)可用下列公式計算： 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中間變量多于兩個的情況的情況.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= =都在點都在點),(yx具有對具有對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 = =在對應(yīng)在對應(yīng)點點),(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= =在對應(yīng)點在對應(yīng)點),(yx的兩個偏的兩個偏導(dǎo)數(shù)存在，且可用

3、下列公式計算導(dǎo)數(shù)存在，且可用下列公式計算 ;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 證第一個公式證第一個公式: ,uxx yx y = = ,vxx yx yy yy y = = 由于由于 Z = f ( u, v ) 在點在點 ( 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 = =都在點都在點),(yx具有對具有對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= =在對應(yīng)點在對應(yīng)點),(yx的兩個的兩個偏導(dǎo)數(shù)存在，且可用下列公式計算偏導(dǎo)數(shù)存在，且可用下列公式計算 , 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對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= =時時，有有dyyzdxxzdz = =.全微分形式不變形的實質(zhì)全微分形式不變形的實質(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