yxex1的精选

问题详情：曲线y=xex+1在点(0,1)处的切线方程是()A.x-y+1=0           B.2x-y+1=0C.x-y-1=0           D.x-2y+2=0【回答】A.y′=ex+xex,当x=0时,导数值为1,故所求的切线方程是y=x+1,即x-y+1=0....