隐函数求导数y+xy' =e^(x+y) (1+ y')[x-e^(x+y)] y'=e^(x+y)-yx(1-y)y' = y(x-1)y' =[y(x-1)] / [x(1-y)] 请问第二步到第三步是怎么来的?

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隐函数求导数y+xy''=e^(x+y)(1+y'')[x-e^(x+y)]y''=e^(x+y)-yx(1-y)y''=y(x-1)y''=[y(x-1)]/[x(1-y)]请问第二步到第三步是怎么来的?隐函

隐函数求导数y+xy' =e^(x+y) (1+ y')[x-e^(x+y)] y'=e^(x+y)-yx(1-y)y' = y(x-1)y' =[y(x-1)] / [x(1-y)] 请问第二步到第三步是怎么来的?
隐函数求导数
y+xy' =e^(x+y) (1+ y')
[x-e^(x+y)] y'=e^(x+y)-y
x(1-y)y' = y(x-1)
y' =[y(x-1)] / [x(1-y)]
请问第二步到第三步是怎么来的?

隐函数求导数y+xy' =e^(x+y) (1+ y')[x-e^(x+y)] y'=e^(x+y)-yx(1-y)y' = y(x-1)y' =[y(x-1)] / [x(1-y)] 请问第二步到第三步是怎么来的?
隐函数应该是xy=e^(x+y)
把上式代人第二步得(x-xy)y'=xy-y即x(1-y)y' = y(x-1)