Question

一道隐函数求导的题,急!y=1+xe^y 求dy/x=0=什么?

Answer 1

y' = e^y + xe^y*y'

y' = e^y/(1-xe^y)

即 dy/dx = e^y/(1-xe^y) = e^y/(2-y)

x=0时 y=1

代入 即得 dy/dx =e

dy = edx