求z=(x^2+y^2)⼀xy的偏导数

2025-02-01 17:04:04
推荐回答(1个)
回答1:

题目应为:z=(x^2+y^2)/(xy)
偏z/偏x=[xy*2x-(x^2+y^2)y]/(xy)^2
=(2x^2y-x^2y-y^3)/(x^2y^2)
=(x^2y-y^3)/(x^2y^2)
=(x^2-y^2)/(x^2y)
偏z/偏y=-(x^2-y^2)/(xy^2)