(6) f = √|xy|, f<x>(0, 0) = lim<x→0>[f(x,0) - f(0,0)]/x = 0.
(7) u = x^(y^z), u<x> = (y^z)x^(y^z-1), u<x>(1,2,3) = 8.
(8) u = f(xy, xyz), 记 p = xy, q = xyz
∂u/∂x = (∂f/∂p)(∂p/∂x) + (∂f/∂q)(∂q/∂x)
= y(∂f/∂p) + yz(∂f/∂q)
(7) u = x^(y^z), u<x> = (y^z)x^(y^z-1), u<x>(1,2,3) = 8.
(8) u = f(xy, xyz), 记 p = xy, q = xyz
∂u/∂x = (∂f/∂p)(∂p/∂x) + (∂f/∂q)(∂q/∂x)
= y(∂f/∂p) + yz(∂f/∂q)
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第1个回答 2018-06-15
6、fx(0,0)=lim[y→0]【f(0,y)-f(0,0)】/(y-0)=0
7、因为对x求偏导，故把y^z看做常数
故Ux=y^z ·x^(y^z - 1)
故Ux(1,2,3)=2^3 · 1^(2^3 -1)=8
8、=f'1 · x +f'2 · xz
=xf'1+xzf'2
7、因为对x求偏导，故把y^z看做常数
故Ux=y^z ·x^(y^z - 1)
故Ux(1,2,3)=2^3 · 1^(2^3 -1)=8
8、=f'1 · x +f'2 · xz
=xf'1+xzf'2
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