Question

求函数y=e的x次方乘x+2在（0,2)处的切线方程和法线方程

Answer 1

y=xe^x+2 ,
y ' = e^x + xe^x ,令 x = 0 得 k = y ‘ = 1 ,
所以切线方程为 y -2 = 1*(x-0) ,即 x-y+2 = 0 ,
法线方程为 y - 2 = -1*(x-0) ,即 x+y-2 = 0 .