例5设u?f(x,y,z),x?rsin?cos?,y?rsin?sin?,z?rcos?,且x?u?x?y?u?y?z?u?z?0,证明：u与r无关.证因为?u?r??u?x?u?y?u?z??????x?r?y?r?z?r?u?x?sin?sin??u?y?cos??u?z?sin?cos??1??u?u?u?x?y?z???0,r??x?y?z? 所以u与r无关.例6要条件是u证?u?x?y2
?u?u?.?x?y设u(x,y)的二阶偏导数存在且u?0,证明：u(x,y)?f(x)?g(y)的充?必要性.若u(x,y)?f(x)?g(y),则?u?x?u?y?u?x?y2?f?(x)g(y),?f(x)g?(y),?f?(x)g?(y),从而u?u?x?y2?f(x)g(y)?f?(x)g?(y)?f?(x)g(y)?f(x)g?(y)??u?x?y2?u?u?.?x?y充分性.若u??u?u?,即?x?yu???u??u?u?,????y??x??x?y亦即
u???u??u?u?????y??x??x?yu2?0,或??u????x????0,?y?u?从而有???lnu????0,?y??x?所以?lnu?x与y无关,即?lnu?x故lnu???(x)dx??(y),所以?(x)dx?(y)u?e??e?f(x)?g(y).x?y?z??(x),
?例7解设u?xyze注意到,求?pp?q?rqzr?x?y?z,其中p,q,r?Z.u?xe?ye?ze,有?pp?q?rqxyzzr?x?y?z由Leibniz公式,d同理ppp?dppdx(xe)?xdqqdy(ye)?ydrrdz(ze).zdx(xe)?x?Ck?0kp(x)(k)(e)x(p?k)?xe?pe?(x?p)e,xxxddyqq(ye)?(y?q)e,yy
(ze)?(z?r)e,zzd所以?pp?q?rqrrdzzr?x?y?z?(x?p)e?(y?q)e?(z?r)e?(x?p)(y?q)(z?r)exyzx?y?z.
例8??(x,y)?fyy??(x,y),f(x,2x)?x2,设f(x,y)具有二阶连续偏导数,且fxxfx?(x,2x)?x,求fxx??(x,2x),fxy??(x,2x).解因f(x,2x)?x2,两边对x求导,得fx?(x,2x)?2fy?(x,2x)?2x,两边再对x求导,得fxx??(x,2x)?2fxy??(x,2x)?2fyx??(x,2x)?4fyy??(x,2x)?2.从而由fxx??(x,y)?fyy??(x,y)以及f(x,y)有二阶连续偏导数得5fxx??(x,2x)?4fxy??(x,2x)?2又因fx?(x,2x)?x,两边对x求导,得fxx??(x,2x)?2fxy??(x,2x)?1(1)(2)解得f1xx??(x,2x)?0,fxy??(x,2x)?2.3例9已知f2xx(x,y)?xy?x,fy(x,y)?3?y,求f(x,y).解因fx(x,y)?x2y?x,故f(x,y)??(x2y?x)dx?13x3y?12x2??(y),两边对y求导得fx3y(x,y)?3???(y),3与已知条件fy(x,y)?x3?y比较得??(y)?y,积分得?(y)?12y2?C.
所以f(x,y)?13x3y?12x2?12y2?C.
题型三 隐函数的偏导数运算
(1)(2)由
例10解设z?z(x,y)是由方程yze2x?y2x?y?sin(xyz)?0确定的函数,求dz.记F(x,y,z)?yze?sin(xyz),则2x?yFx?(x,y,z)?ye?yzcos(xyz),?yze2x?yFy?(x,y,z)?2yzeFz?(x,y,z)?ye故?z?x?z?y所以dz??z?xdx??z?ydy2x?y2x?y?xzcos(xyz),x?y?xycos(xyz),??Fx?Fz?Fy?Fz??yzcos(xyz)?yeye2x?y2x?y?xycos(xyz)2x?y,?yze2x?y???xzcos(xyz)?2yzeyex?y?xycos(xyz),?[yzcos(xyz)?ye]dx?[xzcos(xyz)?2yzeye2x?yx?y?yze2x?y]dy?xycos(xyz).
2例11设z?z(x,y)由方程F(x?z,y?z)?12(x?y?z)?222确定,且F(u,v)可微,求解?z?x?y,?z和dz.方程两边对x求导?z??z1??z??F1???1???2x?2z??F2????0,?x??x2??x??解得?z?x?x?F1?F1??F2??z.
再方程两边对y求导F1??解得?z?y所以dz??z?xdx??z?ydy?(x?F1?)dx?(y?F2?)dyF1??F2??z.?y?F2?F1??F2??z.?z?1??z???F2???1??2y?2z?2???0,?y?y?y?????z
xyxx?zsint例12设u?f(x,y,z)具有一阶连续偏导数,且e?xy?2,e??0求dudx.解：由题意可知,y?y(x),z?z(x),故dudx??ff?x???y?dydx??fdz?z?dx.对exy?xy?2两边关于x求导,得exy???y?xdy??dy?dx?????y?xdx???0,故dyd??yxx.对ex??x?zsint0tdt两边关于x求导,得ex?sin(x?z)?dzx?z???1??dx??,故dzxdx?1?e(x?z)sin(x?z).将(2)(3)代入(1)得du?fy?f?ex(dx??x?x?y?x?z)??f?1??sin(x?z)?.??ztdt,(1)(2)(3)