第十八章 隐函数定理及其应用
§1 隐函数
1. 方程cosx?siny?exy能否在原点的某邻域内确定隐函数y?f?x?或x?g?y??
分析:隐函数是否存在只须验证题目是否满足隐函数存在定理的条件.
解 令F?x,y??cosx?siny?exy,则有 (1) F?x,y?在原点的某邻域内连续; (2) F?0,0??0;
(3) Fx??sinx?yexy,Fy?cosy?xexy均在原点的上述邻域内连续; (4) Fy?0,0??1?0,Fx?0,0??0.
故由隐函数存在定理知,方程cosx?siny?e在原点的某邻域内能确定隐函数y?f?x?.
xy2. 方程xy?zlny?exz?1在点?0,1,1?的某邻域内能否确定出某一个变量为另外两个变
量的函数?
分析: 本题的解题思路与1题一样.
解 令F?x,y,z??xy?zlny?exz?1,则
(1) F?x,y,z?在点?0,1,1?的某邻域内连续; (2) F?0,1,1??0; (3) Fx?y?ze,Fy?x?xzz,Fz?lny?xexz均在原点的上述邻域内连续; y(4) Fx?0,1,1??2?0,Fy?0,1,1??1?0,Fz?0,1,1??0. 故由隐函数存在定理知,方程xy?zlny?exz?1在点?0,1,1?的某邻域内能确定隐函数
x?f?y,z?和y?g?x,z?.
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3. 求出下列方程所确定的隐函数的导数: (1) x2y?3x4y3?4?0,求
dyydy22; (2) lnx?y?arctan,求; dxxdx22u(3)
e?xyx?a2?y2?a?0?,?2z?e?0,求zx,zy; (4) a?a?y?ye,u?azdyd2y,2; (5) x2?y2?z2?2x?2y?4z?5?0,求zx,zy; 求
dxdx(6) z?f?x?y?z,xyz?,求
?z?x?y,. ?x?y?z分析: 求隐函数的导数(偏导数)通常有三种方法:①用隐函数求导公式;②对所给方程(组) 两边直接求导(偏导数);③用全微分.另一种方法是将隐函数显化(如果可能而且又方便的话),但一般来说这种方法是不行的,只有在特殊条件下才可能使用. 解 (1) 解法1 令F?x,y??xy?3xy?4,则
243Fxdy2y?12x2y3Fx?2xy?12xy,Fy?x?9xy,所以????.
dxFyx?9x3y233242解法2 方程两边对x求导,得2xy?x2dydy?12x3y3?9x4y2?0, dxdxdy2y?12x2y3解得. ??32dxx?9xy(2) 解法1令F?x,y??lnx?y?arctan22yxyx?y,则Fx?2, ??22222xx?yx?yx?yFy?Fxx?yyxy?xdy??,所以???dxFyx?yx2?y2x2?y2x2?y22x?2y?x?y?.
1x2dy1dx?解法2 方程两边对x求导,得
x2?y22x2?y2?y?1????x?dy?ydx,整理得 x2 2
x?ydydxxdy?yx2?y2?dxdyx?yx2?y2,所以dx?x?y?x?y?.
解法3 方程两边分别微分,得
xdx?ydyxdy?ydxx2?y2?x2?y2,解得dyx?ydx?x?y?x?y?.
(3) 解法1设F?x,y,z??e?xy?2z?ez,则Fxyx??ye?,Fxyy??xe?,Fz??2?ez,
Fxye?xyFyxe?xy所以zx??F?z?2;zy??F?z. zeze?2解法2 方程两边分别对x,y求偏导,得:
xy?ye?xy?2zzx?ezx?0,?xe?xy?2zz?ye?y?ezy?0,所以zxez?2;zxe?xyy?ez?2. 解法3 方程两边微分,得
e?xy??ydx?xdy??2dz?ezdz?0,即?ez?2?dz?ye?xydx?xe?xydy,所以
?ye?xyxe?xyye?xyxe?xydzez?2dx?ez?2dy,由全微分公式得zx?ez?2;zy?ez?2.
x?a2?y2(4) 令F?x,y??a?a2?y2?yea,则
Fyeu,Fy??y?x??y???y2?eu?yeuaa2aa2?y2,于是 dydx??FxF?y2?ay???a?a2?y2?a?a2?y2?y?a?a2?y2???ya?y2,
a2?y2?ay?a2?y2a2?y2dyyd2yd?dydx?ydya2?y2dxa2ydx2?dx???dx????a2?y2??a2?y2?2.
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(5) 令F?x,y,z??x2?y2?2x?2y?4z?5,则Fx?2?x?1?,Fy2?y?1?,Fz?2?z?2?, 所以zx??Fxx?1y?1. ??,zy??Fzz?2z?2f1?yzf2.
1?f1?xyf2(6) 把z看成x,y的函数,方程两边对x求偏导数,则有
zx?f1??1?zx??f2??yz?xy?zx??zx?把x看成y,z的函数,方程两边对y求偏导数,则有
0?f1??1?xy??f2??yz?xy?xz??xy?1??f1?xyf2?. f1?yzf21?(f1?xyf2).
f1?xzf222把y看成x,z的函数,方程两边对z求偏导数,则有
1?f1??yz?1??f2??xy?xz?yz??yz?dzd2z,2. 4. 设z?x?y,其中y?f?x?为由方程x?xy?y?1所确定的隐函数,求
dxdx22dzdy2x2?y2dy2x?y22?解 由x?xy?y?1得,,又由z?x?y,得. ?2x?2y?dxx?2ydxdxx?2y22??dy?dy??22???4x?yx?2y?2x?y1?2????d2zd?dz?6xdxdx?????4x?2y?故2?. ???22dx?dx?x?2ydx?x?2y??x?2y???5. 设u?x?y?z,其中z?f?x,y?是由方程x?y?z?3xyz所确定的隐函数,
222333求ux及uxx.
Fxx2?yz 解 令F?x,y,z??x?y?z?3xyz,得zx??.于是 ?Fzxy?z2333?zx2?yz2?ux?2x?2z?zx?2??x?xy?z2??,
??
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?zx?x2?2zx?2yzzxxy?z2?zx2?yz2?y?2zzx??2xzy3x3?z3??3xyauxx?2?1? ??2222xy?zxy?z????????????????
6. 求由下列方程所确定的隐函数的偏导数:
(1) x?y?z?e??x?y?z?,求z对于x,y的一阶偏导数和二阶偏导数;
?z?z?2z(2)F?x,x?y,x?y?z??0,求,,和2.
?x?y?x解 (1) 令F?x,y,z??x?y?z?e??x?y?z?,则
Fx?1?e??x?y?z??Fy?Fz?zx?zy??1,zxx?zxy?zyy?0.
(2) 等式两边分别对x,y求偏导数,得
?F1?F2??F2???,. F1?F2?F3?1?zx??0,F2?F3?1?zy??0?zx???1?z??1??y????F3???F3?再将zx对x求偏导数,得
zx2??1?F3?F11?F12?F13?1?zx??F21?F22?F23?1?zx????F1?F2??F31?F32?F33?1?zx??? 2F3 =?122????????FF?2F?F?2F?FFF?F?F?FF33. 3111222123132312F32?? 7. 证明:设方程F?x,y??0所确定的隐函数y?f?x?具有二阶导数,则当Fy?0时,有
FxxFy3?y???FxyFx 证 直接对原方程接连求导两次
FxyFyyFyFxFy. 0Fx?Fy?y??0,?y???
FxFy?Fy?0?.Fxx?111FxFxy?FxFyx?2Fx2Fyy?Fyy???0, FyFyFy5