如何求y=1+xe^y的二阶导数d2y⼀dx2

2024-12-08 21:08:53
推荐回答(2个)
回答1:

y = 1+xe^y...........................................(1)
y' = e^y + xy' e^y..................................(2)
y' = e^y/(1-xe^y)...................................(3)
(2) 式再对x求一次导数:
y'' = y'e^y + y'e^y + xy''e^y + xy'^2e^y
解出:
y'' = y'(2+xy')e^y / (1+xe^y)..................(4)
将(3) y‘ 代入(4) ,整理后即为所求。

回答2: