设y=y(x)由方程y-xe^y=1所确定，则d^y/dx^2|x=0得多少

Question

如题所述

Answer 1

y-xe^y=1
当x=0时,y=1
两边同时对x求导得
dy/dx-e^y-xe^y*dy/dx=0
dy/dx=e^y/(1-xe^y)
dy/dx|(x=0,y=1)=e
d^2y/dx^2
=e^y*dy/dx*(1-xe^y)-e^y*(-e^y-xe^y*dy/dx)/(1-xe^y)^2 |(x=0,y=1)
=[e*e-e*(-e)]/1
=2e^2

Answer 2

y-xe^y=1
y'-e^y-xe^y*y'=0
y'=e^y/(1-xe^y)
y''=(y'e^y(1-xe^y)-(-e^y-xy'e^y)e^y)/(1-xe^y)^2
x=0时，y=1 y'=e^1 /(1-0)=e
y''=(e*e^1 *(1-0)-(-e^1-0)*e^1)/(1-0)^2=e^2+e^2=2e^2

Answer 3

什么意思？是d^2 y / d x^2（当x=0），还是dy/dx（当x=0）？本回答被提问者采纳