Question

两道导数题

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Answer 1

2、x^(2/3)+y^(2/3)=1
上式两边对x求导
(2/3)*x^(-1/3)+(2/3)*y^(-1/3)*y'=0
y^(-1/3)*y'=-x^(-1/3)
y'=-(y/x)^(1/3)

3、y=1+xe^y
两边对x求导
y'=e^y+xe^y*y'
(1-xe^y)*y'=e^y
y'=e^y/(1-xe^y)

追问

能不能求一下二阶导

追答

2、
y'=-(y/x)^(1/3)=-y^(1/3)*x^(-1/3)
y''=-(1/3)*y^(-2/3)*y'*x^(-1/3)+(1/3)*y^(1/3)*x^(-4/3)
=(1/3)*[y^(1/3)*x^(-4/3)+y^(-1/3)*x^(-2/3)]

3、
y'=e^y/(1-xe^y)=(y-1)/x(2-y)
y''=[y'*x(2-y)-(y-1)(2-y)+(y-1)x]/[x(2-y)]^2
=(y-1)(y+x-1)/[x(2-y)]^2