Question

求由方程∫(y到0)e^tdt+∫(x^2到x)1/tdt=0所确定的隐函数的二阶导数，每一步详细说吗谢谢

Answer 1

求由方程【0，y】∫e^tdt+【x，x²】∫1/tdt=0所确定的隐函数的二阶导数。

解：e^t∣【0，y】+lnt∣【x，x²】=0
即有e^y-1+lnx²-lnx=0
也就是隐函数F(x，y)= e^y+lnx-1=0，求d²y/dx².
dy/dx=y'=-(∂F/∂x)/(∂F/∂y)=-(1/x)/(e^y)=-1/(xe^y)
d²y/dx²=dy'/dx=[e^y+(xe^y)y']/(xe^y)².=(e^y-1)/(xe^y)²

Answer 2

两边同时求导即可得
(e^y)y'+(1/x^2)(x^2)'-(1/x)(x)'=0
y'e^y+1/x=0
y'=-e^(-y)/x
y''=e^(-y) y'/x+e^(-y)/x²=e^(-y)[-e^(-y)/x]/x+e^(-y)/x²=e^(-y)[1-e^(-y)]/x²

Answer 3

∫(0->y)e^tdt+∫(x->x^2)(1/t) dt=0
e^y - e + lnx + C =0
e^y .y' + 1/x =0
y' = -(1/x)e^(-y)
y'' = -( -(1/x)e^(-y). y' -(1/x^2)e^(-y) )
= (1/x)e^(-y) .(-1/x)e^(-y) + (1/x^2) e^(-y)
= (1/x^2)e^(-y) [ -e^(-y) +1 )