→ yhliu :重點是: 緊記 z 是 x, y 的函數. 05/25 10:55
推 caregix :嗯~謝謝QQ 05/25 17:19
※ 引述《caregix (人心歸善)》之銘言：
: 題目:設z+ln(z)=xy,求Zx及Zxy~
: (註:Zx就是Z對x取偏微,Zxy則是Z對x取完偏微再對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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※ 編輯: yhliu 來自: 125.233.156.41 (05/25 10:54)