y=1+x^2e^y的二阶导数 谁给个详细过程谢谢啦

Question

结果麻烦给个化简的结果呗谢谢了

Answer 1

两边对x求导
y'=2xe^y+x^2e^y*y' (*)
y'=2xe^y/(1-x^2e^y)
两边再对(*)求导
y''=2e^y+2xe^y*y'+2xe^y*y'+x^2e^y (y')^2+x^2e^y*y''
y''=(2e^y+2xe^y*y'+2xe^y*y'+x^2e^y (y')^2)/(1-x^2e^y)
=[2e^y+4xe^y*2xe^y/(1-x^2e^y)+x^2e^y*(2xe^y)^2/(1-x^2e^y)^2]/(1-x^2e^y)

Answer 2

y=1+x^2e^y
y'=2xe^y+x^2y'e^y (1-x^2e^y)y'=2xe^y y'=2xe^y/(1-x^2e^y)
y''=2e^y+2xy'e^y+[(x^2e^y)y' ]'
=2e^y+2xy'e^y+(2xy'e^y+x^2y'^2e^y)+x^2e^yy''
(1-x^2e^y)y''=2e^y+xe^y(4y'+xy'^2)
(1-x^2e^y)y''=2e^y+xe^y(8xe^y/(1-x^2e^y)+2x^3e^y/(1-x^2e^y)^2 )
y''=[2e^y+xe^y(8xe^y/(1-x^2e^y)+2x^3e^y/(1-x^2e^y)^2 )]/(1-x^2e^y)

Answer 3

书《

Answer 4

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Answer 5

这不就是隐函数求导吗？有困难？