z=f(x,y) xy+yz+xz=1 ,求dz

学习 时间:2026-03-30 12:11:06 阅读:5126
z=f(x,y) xy+yz+xz=1 ,求dz

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优雅的寒风

心灵美的宝贝

2026-03-30 12:11:06

dz=(∂ z / ∂ x)dx+(∂ z / ∂ y)dy
xy+yz+xz-1=0
设g(x,y,z)=xy+yz+xz-1
  ∂ g / ∂ x =y+z ①
  ∂ g / ∂ y =x+z ②
  ∂ g / ∂ z =x+y ③
∂ z / ∂ x = - ①/③= -(y+z)/(x+y)
∂ z / ∂ y = - ②/③= - (x+z)/(x+y)
所以
d z = -(y+z)d x /(x+y)- (x+z)dy /(x+y)
  = - [(y+z)d x +(x+z)d y ] /(x+y)

最新回答共有2条回答

  • 甜甜的小丸子
    回复
    2026-03-30 12:11:06

    dz=(∂ z / ∂ x)dx+(∂ z / ∂ y)dyxy+yz+xz-1=0设g(x,y,z)=xy+yz+xz-1  ∂ g / ∂ x =y+z ①  ∂ g / ∂ y =x+z ②  ∂ g / ∂ z =x+y ③∂ z / ∂ x = - ①/③= -(y+z)/(x+y)∂ z / ∂ y = - ②/③= - (x+z)/(x+y)所以d z = -(y+z)d x /(x+y)- (x+z)dy /(x+y)  = - [(y+z)d x +(x+z)d y ] /(x+y)

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