曲线y=x^3过点(1,1)处的切线方程是什么?

答：y=x^3
求导：y'(x)=3x^2
点（1,1）在曲线上
设切点为（a,a^3)
则切线斜率k=3a^2=(a^3-1)/(a-1)
3a^3-3a^2=a^3-1
2a^3-3a^2+1=0
(2a^3-2a^2)+1-a^2=0
2(a-1)a^2-(a-1)(a+1)=0
(a-1)(2a^2-a-1)=0
(a-1)(2a+1)(a-1)=0
解得：a=1或者a=-1/2
a=1时：k=3,切线为y-1=3(x-1),y=3x-2
a=-1/2时：k=3/4,切线为y-1=(3/4)(x-1),y=3x/4+1/4