(1)设u??(x),v??(x)在点x处可导,z?f(u,v)在对应于x的点(u,v)处可微,则复合函数z?f[?(x),?(x)]在点x处可导,且dz?u?dudx??z?v?dvdx??zdx?fu????fv???.(2)设u??(x,y),v??(x,y)在点(x,y)处偏导数存在,z?f(u,v)在对应于点(x,y)的点(u,v)处可微,则复合函数z?f[?(x,y),?(x,y)]在点(x,y)处可导,且?z?x??z?u?u??x??z?v??v?x,?z?z?u?z?v?y??u??y??v??y.函数求导法
(1)一个方程确定的隐函数(i)二元方程F(x,y)?0确定的一元隐函数y?y(x)的导数为dydx??FxF.y(ii)三元方程F(x,y,z)?0确定的二元函数z?z(x,y)的偏导数为?z?x??Fx?zFyF,z?y??F.z(2)方程组确定的隐函数(i)由两个方程构成的四元方程组?F(x,y,u,v)?0??G(x,y,u,v)?0确定的两个二元隐函数u?u(x,y),v?v(x,y)的偏导数ux,vx；uy,vy的求法是：分别对方程组关于x,y求导(此时注意u,v是x,y的函数),然后以ux,vx；uy,vy为未知数解二元一次方程组即可.(ii)由两个方程构成的三元方程组?F(x,y,z)?0?确定的两个一元隐函?G(x,y,z)?0数y?y(x),z?z(x)的导数dy的求法是：对方程组关于dx,dzdxx求导,然后以dydzdx,dx为未知数解二元一次方程组即可.一般地,在一定条件下,对于有m个方程、n个变量(n?m)的方程组来说,有m个因变量就有n?m个自变量.关键是事先要明确哪些变量是因变量,哪些变量是自变量,这应根据具体问题确定.例如题目所求的是?z?z?x,?y,即可知z是因变量,x,y是自变量. 二、示例
题型一 偏导数、连续、可微的关系
4. 隐
例1讨论函数?22?(x?y)sinz?f(x,y)???0,?1x?y22,x?y?0x?y?02222在点(0,0)处：(1)是否连续？(2)是否存在偏导数？(3)是否可微？(4)fx(x,y)和fy(x,y)在点(0,0)是否连续？解(1)由有界函数与无穷小的乘积为无穷小得limf(x,y)?lim(x?y)sinx?0y?0x?0y?0221x?y22?0?f(0,0),故f(x,y)在点(0,0)处连续.(2)因?f?x同理(0,0)?limf(x,0)?f(0,0)xxsin?limx?021|x|?limxsinx?01|x|x?0x?0,?f?y(0,0)?0(因f(x,y)?f(y,x)),故f(x,y)在点(0,0)处的偏导数存在.(3)由(2)知fx(0,0)?0,fy(0,0)?0,故?z?(fx(0,0)dx?fy(0,0)dy)??z?((?x)?(?y))sin其中??(?x)?(?y),于是lim即?z?fx(0,0)dx?fy(0,0)dy?o(?),所以f(x,y)在点(0,0)处可微,且dz(4)当(x,y)?(0,0)时,fx(x,y)?2xsinfy(x,y)?2ysin而lim2xsinx?0y?0x?0y?022221(?x)?(?y)22??sin21?,?z?(fx(0,0)dx?fy(0,0)dy)??0??lim?sin??01??0,?0.1x?y1x?y2222?xx?y22cos1x?y22,?yx?y22cos1x?y22,1x?y22?0,又limxx?y22x?0y?xcos1x?y22?limx|x|x?0cos12|x|不存在,故limfx(x,y)不存在,所以fx(x,y)在点(0,0)处不连续.同理fy(x,y)x?0y?0在点(0,0)处不连续.注：由本例可知,fx(x,y),fy(x,y)在点(x,y)处连续是f(x,y)处可微的充分条件而不是必要条件.
例2讨论函数xy?,?22z?f(x,y)??x?y?0,?(x,y)?(0,0)(x,y)?(0,0)在点(0,0)处是否连续？fx(0,0),fy(0,0)是否存在？解选择路径y?kx,则limf(x,y)?limx?0y?kxxyx?y22x?0y?kx?limkx2222x?0x?ky?k1?k2,该极限值随k取值不同而不同,故limf(x,y)x?0y?0不存在,从而f(x,y)在点(0,0)处不连续.又因0?f?x同理?f?y所以fx(0,0),fy(0,0)存在.注：由本例可知,对多元函数而言,可偏导??连续,这与一元函数可导必定连续绝然不同.二元函数z?f(x,y)在点P(x,y)处可微、连续及可偏导之间有如下关系：(0,0)(0,0)?limf(x,0)?f(0,0)xx?0?limxx?02?0x?0,?0,
题型二 显函数的偏导数运算
例3设z?(x?y)e?z?x?arctanyx22?arctanyx,求dz与2?z?x?y2.yy解?2xe?arctan?y??arctanxx?(x?y)????e?(2x?y)e,?22?x???y?1????x?2221?z?y所以?2ye?arctanyx?(x?y)?1?y?1????x?2?1x?e?arctanyx?(2y?x)e?arctanyx,dz?e?z?x?y2?arctanyx?(2x?12y)dx?(2y?x)dy?.?1x?e?arctanyx?e?arctanyx?(2x?y)??y?xy?xx?y2222e?arctanyx.?y?1????x?
12a例4设?,?具有二阶连续偏导数,z?12??(y?ax)??(y?ax)???y?axy?ax?(t)dt,试求?z?x解22?a?z?x2?z?y12a2a2a22.12a?????(y?ax)?a???(y?ax)?a??12??(y?ax)?a??(y?ax)?a????(y?ax)???(y?ax)????(y?ax)??(y?ax)?,????(y?ax)?a????(y?ax)?a?????(y?ax)?a???(y?ax)?a?22?z?x22??121212????(y?ax)????(y?ax)?????(y?ax)???(y?ax)?212a12a2a同理?z?y?z?y所以22????(y?ax)???(y?ax)??????(y?ax)????(y?ax)??2??(y?ax)??(y?ax)?,???(y?ax)???(y?ax)?,
??z?x2?a2?z?y2?0.