设y=y(x)由方程y-xe^y=1所确定,则d^y⼀dx^2|x=0得多少

2025-04-06 19:06:39
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回答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