Question

画红色笔的，谢谢

Answer 1

8. d ( xy + ln y = 1 )
ydx + xdy + 1/y * dy = 0
y' = dy/dx = - y^2 / ( xy + 1 )

9.2. d ( xy = e ^(x+y) )
ydx + xdy = e^(x+y) ( dx + dy )

y' = dy/dx = ( y - e ^(x+y) ) / ( e^(x+y) - x )

追问

大神可否做在纸上给我看看，酱紫看不懂啊

追答

d ( xy + ln y = 1 )
xdy + ydx + 1/y * dy = 0

ydx = - xdy - 1/y dy
ydx = - ( x + 1/y ) dy

两边除以 - ( x + 1/y )
- y / ( x + 1/y ) * dx = dy

两边除以 dx
- y / ( x + 1/y ) = dy/dx

简化
- y^2 / ( xy + 1 ) = dy/dx

y' = dy/dx

所以
y'= dy/dx = - y^2 / ( xy + 1 )