Question

求下列方程所确定的隐函数的导数dy/dx xy=e∧x+y

求下列方程所确定的隐函数的导数dy/dx xy=e∧x+y y=1-xe∧y

Answer 1

方法一
1.两边对x求导
y+xy'=e^x+y'
(x-1)y'=e^x-y
dy/dx=y'=(e^x-y)/(x-1)
2.
两边对x求导
y'=-e^y-xe^y*y'
y'=-e^y/(1+xe^y)

方法二，构建函数F(x,y)=0,dy/dx=-Fx/Fy
1.
F(x,y)=xy-(e^x+y)
Fx=y-e^x
Fy=x-1
dy/dx=-(y-e^x)/(x-1)
2.
F(x,y)=y+xe^y-1
Fx=e^y
Fy=1+xe^y
dy/dx=-e^y/(1+xe^y)

Answer 2

题干不详，无法作答