求Y=1+xe^y的微分麻烦给一下过程,越详细越好,谢谢

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y=1+xe^y ==>y'=(1+xe^y )'
==>y'=(xe^y)'
==>y'=1*e^y+xe^y*y'
==>y'(1-xe^y)=e^y
==>y'=e^y/(1-xe^y)
因为y=1+xe^y,则1-xe^y=2-y,得y'=e^y/(2-y)
即dy/dx=e^y/(2-y)
dy=[e^y/(2-y)]dx
======以下答案可供参考======
供参考答案1:
对x求导y'=x'*e^y+x*(e^y)'
y'=e^y+x*e^y*y'
y'=e^y/(1-xe^y)
所以dy=[e^y/(1-xe^y)]dx