例4设=x+y+x,xy-),俱具有二阶连续偏导数, 求 Ow 3 82 及 解令=x+y+x,v=xy2,则1=f(,y) aw af au of f+y=/2 u Ox OV OX 02y=(f+ye、可+y2+y9 axa az f1+x/2+y2+y2/1+xy2=/2 =升1+y(x+z)f12+y2+xy2=/22 小: af2af2 au. af2 a f21+xyf22 oz ou O V O 百贝贝这回下页 结束首页 上页 返回 下页 结束 铃 引入记号 u f u v f    = ( , ) 1  u v f u v f     = ( , ) 1 2  同理有 2 f   1 1 f   2 2 提示 f  等 11 12 1 1 1 f xyf z v v f z u u f z f = +        +       =    11 12 1 1 1 f xyf z v v f z u u f z f = +        +       =    11 12 1 1 1 f xyf z v v f z u u f z f = +        +       =    21 22 2 2 2 f xyf z v v f z u u f z f =  +        +       =    21 22 2 2 2 f xyf z v v f z u u f z f =  +        +       =    21 22 2 2 2 f xyf z v v f z u u f z f =  +        +       =    解 令u=x+y+z v=xyz 则w=f(u v) 2 2 2 1 1 1 2 2 = f + y(x+ z) f  + yf + xy zf   下页 例4 设w=f(x+y+z xyz) f具有二阶连续偏导数 求 x w   及 x z w    2  1 2 f yzf x v v f x u u f x w = +       +      =    1 2 f yzf x v v f x u u f x w = +       +      =    z f yf yz z f f yzf x z z w    + +    +  =   =    2 2 1 1 2 2 ( ) z f yf yz z f f yzf x z z w    + +    +  =   =    2 2 1 1 2 2 ( ) 22 2 11 12 2 21 = f + xyf  + yf + yzf  + xy zf  22 2 11 12 2 21 = f + xyf  + yf + yzf  + xy zf  22 2 11 12 2 21 = f + xyf  + yf + yzf  + xy zf 