Question

求下列函数的一阶偏导数，求过程

Answer 1

方法如下，
请作参考：

追答

Answer 2

(1) z = u^2lnv, u = y/x, v = 3x-2y
∂z/∂x = (2u∂u/∂x)lnv + (u^2/v)∂v/∂x
= (-2y^2/x^3)ln(3x-2y) + 3y^2/[x^2(3x-2y)]
∂z/∂y = (2u∂u/∂y)lnv + (u^2/v)∂v/∂y
= (2y/x^2)ln(3x-2y) - 2y^2/[x^2(3x-2y)]
(2) u = yz-xlny, u 为 x，y 的函数 ？题目不明确。
两边分别对 x，y 求偏导数， 得
∂u/∂x = y∂z/∂x - lny ， ∂z/∂x = (∂u/∂x+lny)/y ;
∂u/∂y = z+y∂z/∂y - x/y, ∂z/∂y = (∂u/∂y+x/y-z)/y.
(3) z = e^(uv), u = x^2+y^2, v = 3y/x - 2x/y
∂z/∂x = e^(uv)(v∂u/∂x+u∂v/∂x)
= e^[(x^2+y^2)(3y/x - 2x/y)][2(3y-2x^2/y)-(x^2+y^2)(3y/x^2+2/y)]
∂z/∂y = e^(uv)(v∂u/∂y+u∂v/∂y)
= e^[(x^2+y^2)(3y/x - 2x/y)][2(3y^2/x-2x)+(x^2+y^2)(3/x+2x/y^2)]