==Δx-2.
由导数的定义,得f′(-1)= (Δx-2)=-2.
所以f′(1)·f′(-1)=×(-2)=-1.
8.解:设点(1,1)处的切线斜率为k,则
k=f′(1)=
=
= [3+3Δx+(Δx)2]=3,
∴点(1,1)处的切线方程为y-1=3(x-1),
即3x-y-2=0.