1.已知f(x)=x2+2xf′(2014)+2014lnx,则f′(2014)=( )
A. 2015 B. -2015
C. 2014 D. -2014
解析:f′(x)=x+2f′(2014)+,所以f′(2014)=2014+2f′(2014)+,即f′(2014)=-(2014+1)=-2015.
答案:B
2.曲线y=在点(1,-1)处的切线方程为( )
A. y=x-2 B. y=-3x+2
C. y=2x-3 D. y=-2x+1
解析:由题意得y=1+,所以y′=,所以所求曲线在点(1,-1)处的切线的斜率为-2,故由直线的点斜式方程得所求切线方程为y+1=-2(x-1),即y=-2x+1.
答案:D