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'
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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)-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?
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'](/uploads/image/z/18463984-16-4.jpg?t=%E9%9A%90%E5%87%BD%E6%95%B0%E6%B1%82%E5%AF%BC%E6%95%B0y%2Bxy%27+%3De%5E%28x%2By%29+%281%2B+y%27%29%5Bx-e%5E%28x%2By%29%5D+y%27%3De%5E%28x%2By%29-yx%281-y%29y%27)
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)
把上式代人第二步得(x-xy)y'=xy-y即x(1-y)y' = y(x-1)