yxex1的精選

問題詳情：曲線y=xex+1在點(0,1)處的切線方程是()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,當x=0時,導數值為1,故所求的切線方程是y=x+1,即x-y+1=0....