Question

全微分z=(x^2+y^2)e^[(x^2+y^2)/xy] 最后有过程，可以的话偏导数是多少也说一下？

Answer 1

记：z = f(x，y) = (x^2+y^2)e^[(x^2+y^2)/xy] = u(x，y) e^[u(x，y)/v(x,y)]
其中： u(x，y) = (x^2+y^2)；
v(x，y) = xy；
全微分： dz = df(x，y) = [∂f(x，y)/∂x] dx + [∂f(x，y)/∂y] dy
∂f(x，y)/∂x = ∂u(x，y)/∂x e^(u/v) + u e^(u/v) ∂(u/v)/∂x
∂f(x，y)/∂y = ∂u(x，y)/∂y e^(u/v) + u e^(u/v) ∂(u/v)/∂y
∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^2+y^2) (2x-(x^2+y^2))/(x^2y)]
∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (x^2+y^2) (2y-(x^2+y^2))/(xy^2)]
这是 z对x，y的两个偏导数。
z的全微分：
dz = df(x，y) = e^[(x^2+y^2)/xy] {[2x + (x^2+y^2) (2x-(x^2+y^2))/(x^2y)] dx +
+ [2y + (x^2+y^2) (2y-(x^2+y^2))/(xy^2)] dy }

追问

∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^2+y^2) (2x-(x^2+y^2))/(x^2y)]
∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (x^2+y^2) (2y-(x^2+y^2))/(xy^2)]
不应该是
∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^2+y^2) (2x^2-(x^2+y^2))/(x^2y)] 吗？
∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (x^2+y^2) (2y^2-(x^2+y^2))/(xy^2)]

追答

我检查了一下，您说的对，应该是：
∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^4 - y^4)/(x^2y)]
∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (y^4 - x^4)/(xy^2)]
dz = e^[(x^2+y^2)/xy] {[2x + (x^4 - y^4)/(x^2y)] dx + [2y + (y^4 - x^4)/(xy^2)] dy }