求由方程e^y+xy-e=0,所确定的隐函数的导数dy⼀dx.

2024-11-05 02:36:14
推荐回答(4个)
回答1:

两边对x求导,注意y是关于x的函数:
(e^y)'*y'+(xy)'=0
e^y*dx/dy+y+x*dy/dx=0
解得:dy/dx=-y/(e^y+x)
由原方程得:x=(e-e^y)/y
代入得:dy/dx=-y^2/[(y-1)e^y+e]
注意:最后尽量化为y的形式,不要有x出现.我大一学微积分时老师强调的.

回答2:

求导后得e^ydy+ydx+xdy=0,再同时除以dx,得e^ydy/dx+y+xdy/dx=0,即dy/dx=-y/(e^y+x)

回答3:

解:对方程两对求导,得
(e^y)*y'+y+xy'=0
整理得y'=-y/(x+e^y)
所以dy/dx=-y/(x+e^y)

回答4:

e^y+xy-e=0

d(e^y) + d(xy) - d(e) = 0

e^y dy + xdy + ydx = 0

(e^y + x)dy = -ydx

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