Question

e^(xy)+x+y=2隐函数求二阶导

太难算啦啦啦啦啦！（谁出的题！）
注意是求二阶导，不是求一阶导

Answer 1

一阶导和二阶导就是按概念算就行
两边同时对x求导得到
(xy'+y)e^(xy) +1+y'=0
y' =-( ye^(xy)+1)/(1+xe^(xy))
y'' =- [( ye^(xy)+1)'(1+xe^(xy)) - ( ye^(xy)+1)(1+xe^(xy))']/(1+xe^(xy))^2
然后展开算即可

Answer 2

e^(xy)+x+y = 2， 两边对 x 求导，由复合函数求导法则，得
e^(xy)(y+xy') + 1 + y' = 0, (1)
得 y' = -[1+ye^(xy)]/[1+xe^(xy)]
式 (1) 两边再对 x 求导
e^(xy)(y+xy')^2 + e^(xy)(2y'+xy'') + y'' = 0
y'' = -e^(xy)[2y'+(y+xy')^2]/[1+xe^(xy)]
= -e^(xy)[y^2+(2+2xy)y'+x^2y'^2]/[1+xe^(xy)]
= -e^(xy){y^2-(2+2xy)[1+ye^(xy)]/[1+xe^(xy)]
+x^2[1+ye^(xy)]^2/[1+xe^(xy)]^2}/[1+xe^(xy)]
= -e^(xy){y^2[1+xe^(xy)]^2-(2+2xy)[1+ye^(xy)][1+xe^(xy)]
+x^2[1+ye^(xy)]^2}/[1+xe^(xy)]^3