3.设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f为可微函数,验证-|||-dfrac (1)(x)dfrac (partial z)(partial x)+dfrac (1)(y)dfrac (partial z)(partial y)=dfrac (z)({y)^2},

参考答案与解析:

相关试题

7.设 =dfrac (y)(f({x)^2-(y)^2)} 其中f为可导函数,验证: dfrac (1)(x)dfrac (partial z)(partial x)+dfrac (1)(y)dfr

7.设 =dfrac (y)(f({x)^2-(y)^2)} 其中f为可导函数,验证: dfrac (1)(x)dfrac (partial z)(partia

  • 查看答案
  • 设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f(u)为可导函数,验证-|||-.dfrac (1)(x)dfrac (dz)(partial x)+dfrac (1)(y)dfra

    设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f(u)为可导函数,验证-|||-.dfrac (1)(x)dfrac (dz)(partial

  • 查看答案
  • 设 =f(xy,(x)^2+(y)^2), 其中 f 可微,则 dfrac (partial z)(partial x)= __

    设 =f(xy,(x)^2+(y)^2), 其中 f 可微,则 dfrac (partial z)(partial x)= __

  • 查看答案
  • 设=dfrac (y)(f({x)^2-(y)^2)},其中f为可导函数.验证=dfrac (y)(f({x)^2-(y)^2)}.

    设=dfrac (y)(f({x)^2-(y)^2)},其中f为可导函数.验证=dfrac (y)(f({x)^2-(y)^2)}.设,其中f为可导函数.验证.

  • 查看答案
  • 5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(partial y)=1.

    5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(pa

  • 查看答案
  • 2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac (partial f)(partial

    2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac

  • 查看答案
  • sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial y)= () 。

    sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial

  • 查看答案
  • 3.设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(partial y).

    3.设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(part

  • 查看答案
  • 3.设 +2y+z-2sqrt (xyz)=0 ,求 dfrac (partial z)(partial x) 及 dfrac (partial z)(partial y) .

    3.设 +2y+z-2sqrt (xyz)=0 ,求 dfrac (partial z)(partial x) 及 dfrac (partial z)(part

  • 查看答案
  • 殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A dfrac (y)({x)^2}[ 2

    殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A

  • 查看答案