Question

高数微分计算？

Answer 1

e^(x+y)+xy=1,两边同时求导,得e^(x+y)（1+y'）+y+xy=0,所以dy=（y-e^(x+y))/(e^(x+y)+x)dx

Answer 2

e^(x+y) -xy =1
(dx+dy).e^(x+y) -( xdy +ydx ) =0
[e^(x+y) -x ] dy = [y- e^(x+y) ] dx
dy = { [y- e^(x+y) ]/[e^(x+y) -x ] } dx

Answer 3

e^(x+y) - xy = 1
de^(x+y) - d(xy) = 0
e^(x+y)(dx+dy) - ydx - xdy = 0
dy = [y-e^(x+y)]dx/[e^(x+y)-x]

Answer 4

就是方程两边关于x求导得(1+y')e^(x+y)-y-xy'=1. 所以y'=[1+y-e^(x+y)]/(e^(x+y)-x). 所以dy=[1+y-e^(x+y)]/(e^(x+y)-x)·dx.

Answer 5