(1）设u=f(x,y,z)，f具有一阶连续偏导数，而z=z(x,y)是由方程Qu Qux2+y2+z+e°=1确定的隐函数，求ax'ay2x解x2+y2 +z+e =1, 2x+zx+e'z, =0, zxee1+e-+.-1-12yf'2y2y+z, +e'z, =0, z,u, = f'+ f'z, = f'L1+e*1+eOz Oz(2) 设z=u lnv, 而u=, v= 3x-2y, 求ax'ay13x22xln(3x - 2y)12zx = 2uln vJ2y(3x - 2y)V2x22xln(3x - 2y)(-2) =z, = 2ulnv.(J3y(3x - 2y)000811个不不高尊数学教学部不不
高等数学教学部 11 1, 2 2 z x y z e 2 0, x z x x z e z ; 1 2 x z e x z 2 0, y z y y z e z ; 1 2 y z e y z x x u f f z 1 3 , 1 2 3 1 z e xf f y y u f f z 2 3 . 1 2 3 2 z e yf f 3 1 2 ln 2 v u y z u v x , (3 2 ) 2 ln(3 2 ) 3 2 2 2 y x y x y x x y ) ( 2) 1 2 ln ( 2 2 v u y z u v y . (3 2 ) 2 ln(3 2 ) 2 2 2 3 y x y x y x x y
S(3) 若x2 + y2 + z2 - 4z = 0, 求dz;xdx + ydyx2 + y2 + z2 - 4z = 0, 2xdx +2ydy + 2zdz - 4dz = 0, dz =2-z80，求(4) 设函数z=z(x,J)由方程f(x+ y+z2)=z确定,ff'.(1+2zx) =zx, zx =21-2zflGea'z中(5）设 z=f(x+y,xy)，其中f具有二阶连续偏导数，求ax'axdyzx = f'+ yf",zxy =(f" +xfi")+ f'+ y(f2 +xf) = f" +(x+ y)fi" +xyf + f)提示 z,=f'+xf’,f = f(x+ y,xy), f'= f'(x+ y,xy), f'= f'(x+ y,xy)000812个不高等数学教学部不不不
高等数学教学部 12 4 0, 2 2 2 x y z z 2xdx 2 ydy 2zdz 4dz 0, . 2 z xdx ydy dz (1 2 ) , x x f zz z . 1 2zf f z x , 1 2 z f yf x ( ) ( ) 11 12 2 21 22 z f xf f y f xf xy ( ) . 11 12 22 2 f x y f xyf f