y?xy?1xyxy??????ylnx?ylnxln?lnx?； ??lnx??yln(lnx)??lnx?? （3）
?zxyxy???lnx?.ln?lnx??xy?y?x?lnx?.ln?lnx?. ?y???z?x2yx.fx2?y2,xy?x2y.fx2?y2,xyx ?x???????? ?2xy.f?x2?y2,xy??x2y.?f1?.2x?f2?.y?；
???z?x2yy.fx2?y2,xy?x2y.fx2?y2,xyy ?y???????? ?x2.f?x2?y2,xy??x2y.?f1?.2y?f2?.x?.
?z?z?y??z?xy. 2.设z?xy?x???，其中??u?是可微函数，证明： x?y?y?x?x?5.设u?f?x,y,z??ex?y12?z2，而z?x2siny，求
?u?u，. ?y?x?2z?2z?2z6.求下列函数的2，和2.
?x?y?y?x（1）z?f?xy,y?；（2）z?f?sinx,cosy,ex?y?. 【解】
（1）由z?f?xy,y?得
?z?z?xf1??f2?； ?yf1?，?y?x?2z?????y2f11??； 2?y?f1??x?y?yf11?x?2z????f12????f1??xyf11???yf12??； ?f1??y?f1??y?f1??y?xf11
?x?y?2z??2???????????????? ?????xf?fy12y?x?xf11?f12???xf21?f22??xf11?2xf12?f22. 2?y?2z【注意：书中有关2的答案有误】.
?y（2）由z?f?sinx,cosy,ex?y?得
?z?z??siny.f2??ex?yf3?； ?coxs.f1??ex?yf3?；?y?x??2z?x.f1??x?ex?yf3?x 2??cos?x?????ex?yf13??? ??sinx.f1??cosx?cosxf11???ex?yf33??? ?ex?yf3??ex?y?cosx.f31??????2z???cosx.f1??y?ex?yf3?y
?x?y?????ex?yf13????ex?yf3??ex?y?sinyf32???ex?yf33??； ?cosx??sinyf12???cosxex?yf13???ex?yf3??sinyex?yf32???e2x?2yf33??； ??cosxsinyf12????2z?x?y?? ????siny.f?efy23y
?y2?????ex?yf23??? ??cosy.f2??siny??sinyf22???ex?yf33??? ?ex?yf3??ex?y??sinyf32???2sinyex?yf23????ex?yf3??e2x?2yf33??. ??cosy.f2??sin2yf22???2z【注意：书中有关2的答案有误】.
?ydz. dx【解】①式两端对x求导并注意到z是关于x的函数，得 8.设z?f?x???z?? ①，其中f,?可导，求
dz?dz???f??x???z???x???z??x?f??x???z???1????z?.?
dx?dx? ?f??x???z??????z?.f??x???z??.由②式解得
dzf??x???z???.
?dx1???z?f?x???z??dz. ② dx?2z?z?z9.设z?z?x,y?由方程z?lnz??edt?0 ①得到，求，，.
y?x?y?x?yx?t2【解】（一）①式两端对x求导并注意到z是关于x,y的二元函数得
2?z1?z??e?x?0，即 ?xz?x2?1??z ?1???e?x . ②
?z??x由②式解得
?z?x?z1?ze?x2. （二）①式两端对y求导并注意到z是关于x,y的二元函数得
?z?y?1?zz?y?e?y2?0，即 ??1??z?y2?1?z???y??e . 由④ 式解得 ?z?y??z1?ze?y2. （三）由③式得
? ?2z?z??x2?x?y???1?z??ye???1.?z??x2??1?z?2?y?e【代入④】? ?1?z?y2??x2?1?z?2.???1?z.e??e ??z.e?x2?y2?1?z?3.
10.设f可微，试验证： （1）z??y?1?fx2?y2 ① 满足方程zx?x?1?zy?y?zy2； 【证】?z?x?y??1???x?f?x2?y2????y???1?f2?x2?y2??f?x2?y2????x? ③ ④ ⑤ ??y?f2?x2?y2??f??x2?y2.x2?y2?????x???
???2xyf2x2?y2?f??x2?y2?； ? ?z?1?y1??y???y.f?x2?y2????f?x2?y2??y??1??f?x2?y2??y? ?1?1?2f?x2?y2??y???f2?x2?y2???f??x?y2??.x2?y2??y?????? ?1f?x?y???2y2 f2x2?y?f??x2?y2222?. 所以
1?z1?x?x?zy?y
?1?x???2xyf2?x2?y2?f??x?y?????1?y?1?f?x2?y2??2y222f2?x2?y2?f??x2?y2???? ?1y.1f?x2?y2?【由①式】?1y.zy?zy2. 22（2）z?f?x,y?满足方程???z???z???x???????y???z?z???s.?t，其中x?s?t,y?s?t. 【证】
?z?z?x?z?y?z?z?s??x.?s??y.?s??x??y； ?z?z?x?z?t??x.?t??y.?y?t??z?x??z?y. 22故 ?z?s.?z?t?????z??x??z??y???.????z?z???z???z???x??y???????x???????y???. 14.设函数f?x,y?具有二阶连续偏导数，且满足等式
4?2u?2u?x?12?x?y?5?2u2?y2?0. ①
?2u?0. 试确定a,b的值，使等式在变换??x?ay,??x?by下化为
????【解】因为
?u???u???u?u?u?u ； ?.?.?.1?.1???x???x???x?????????u
?u?u???u???u?u?u?u?.?.?.a?.b?a?b. ?y???y???y?????????x 故有
?2u??u????u??????? ????2?x????x??????2u???2u?????2u???2u????????2.?x?????.?x?????????.?x???2.?x?? ?????2u?2u?2u?2? ?. ② 22?????????2u??u????u? ????????????x?y?????y???y??2u???2u?????2u???2u????????2.?y?????.?y?????????.?y???2.?y?? ?????2u?2u?2u ?a.2??a?b??b.2. ③
????????
??u????u??u?????a?b????2?y????y????22?y??2u???2u?????2u???2u????a????2.?y?????.?y???b??????.?y???2.?y?? ????2?2u?2u2?u ?a. ④ ?2ab?b22????????将②、③、④代入①式左边，得
??2u??2u?2u?2u??2u?2u? ①左?4????2?2???????2???12??a.??2??a?b??????b??2??
????2?2?2u?2u2?u? ?5??a??2?2ab?????b??2?? ??2?2u?2u2?u ?4?12a?5a ??8?12a?12b?10ab??4?12b?5b22?????????2???因此方程①化为
?2?2u?2u2?u4?12a?5a??8?12a?12b?10ab??4?12b?5b?0. ⑤ 22????????2???
郑州大学高等数学下课后习题答案解析
