z=f(x,y) xy+yz+xz=1 ,求dz?
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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),1,把z看做x,y的函数,方程xy+yz+xz=1两边分别对X求偏导数,得y+yz‘(x)+z+xz’(x)=0,其中z‘(x)表示z对x的偏导数。z’(x)=-(y+z)/(x+y)。同理z‘(y)=-(x+z)/(x+y)。所以dz=z’(x)dx+z‘(y)dy=-(y+f(x,y))/(x+y)dx-(x+f(x,y))/(x+y)dy。,2,
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),1,把z看做x,y的函数,方程xy+yz+xz=1两边分别对X求偏导数,得y+yz‘(x)+z+xz’(x)=0,其中z‘(x)表示z对x的偏导数。z’(x)=-(y+z)/(x+y)。同理z‘(y)=-(x+z)/(x+y)。所以dz=z’(x)dx+z‘(y)dy=-(y+f(x,y))/(x+y)dx-(x+f(x,y))/(x+y)dy。,2,
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