求解函数的一阶二阶偏导

2026年09月27日 07:07
有2个网友回答
网友(1):

如图所示,望采纳!

网友(2):

(1) z = ln(x+y^2), ∂z/∂x = 1/(x+y^2), ∂z/∂y = 2y/(x+y^2),
∂^2z/∂x^2 = -1/(x+y^2)^2
∂^2z/∂x∂y = -2y/(x+y^2)^2
∂^2z/∂y^2 = 2(x-y^2)/(x+y^2)^2
(2) z = x^y, ∂z/∂x = yx^(y-1), ∂z/∂y = x^ylnx,
∂^2z/∂x^2 = y(y-1)x^(y-2)
∂^2z/∂x∂y = x^(y-1)+yx^(y-1)lnx
∂^2z/∂y^2 = x^y(lnx)^2