an equation of the line tangent to the graph of f(x)=x(1-2x)^3 at the point (1,-1) is?

题目

an equation of the line tangent to the graph of f(x)=x(1-2x)^3 at the point (1,-1) is?
A.y= -7x+6
B.y= -6x+5
C.y= -2x+1
D.y= 2x-3
E.y= 7x-8

答案

先翻译成中文：
求函数f(x)=x(1-2x)^3在点（1,-1）处的切线方程.
f'(x)=(1-2x)^3+3x(1-2x)^2*(-2)=(1-2x)^2*[(1-8x)]
当x=1,f'(x)=-7
所以切线方程为y+1=f'(x)(x-1)=-7(x-1),化简得y= -7x+6
所以选A.