作者yhliu (老怪物)
看板Math
标题Re: [微积] 偏微分的chain rule问题
时间Wed May 25 10:31:50 2011
※ 引述《caregix (人心归善)》之铭言：
: 题目:设z+ln(z)=xy,求Zx及Zxy~
: (注:Zx就是Z对x取偏微,Zxy则是Z对x取完偏微再对y取偏微)
Dx(z) +(1/z)Dx(z) = y
==> Dx(z) = y/(1+1/z) = yz/(z+1)
Dy(z) +(1/z)Dy(z) = x
==> Dy(z) = xz/(z+1)
Dxy(z) = Dy(Dx(z))
= z/(z+1)+y/(z+1)^2 Dy(z)
= z/(z+1)+y/(z+1)^2 xz/(z+1)
= z/(z+1) +xyz/(z+1)^3
令 u=z+ln(z), 解: z=f(u).
f'(u) =dz/du = 1/(du/dz) = 1/(1+1/z) = z/(z+1)
f"(u) = (d/du)f'(u) = (d/du)[f(u)/(f(u)+1)]
= f'(u)/(f(u)+1)^2 = [z/(z+1)]/(z+1)^2
= z/(z+1)^3
故 z = f(xy),
Dx(z) = f'(u)Dx(u) = f'(xy)y
Dy(z) = f'(u)Dy(u) = f'(xy)x
Dxy(z) = f'(xy)+f"(xy)xy
= z/(z+1) + xyz/(z+1)^3
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◆ From: 125.233.156.41
※ 编辑: yhliu 来自: 125.233.156.41 (05/25 10:54)
1F：→ yhliu :重点是: 紧记 z 是 x, y 的函数. 05/25 10:55
2F：推 caregix :嗯~谢谢QQ 05/25 17:19