y'=e^x+x*e^x.在(0,1)处y'=1.也就是切线的斜率是1.则切线方程为:y-1=1*(x-0)....y=x+1....................如果你没有学过导数,和我说一下.
两边对x求导得3x^2+y'+e^y+xe^y*y'=0把点(0,1)代入得y'+e=0y'=-e所以切线方程是y-1=-ex即ex+y-1=0
