题目内容
A.x+y-2=0 B.x-y+2=0
C.x+y+2=0 D.x-y-2=0
解析:y′=-,∴y′|x=1=-1.
∴切线方程为y-1=-1·(x-1),即x+y-2=0.
答案:A
y=在点A(1,1)处的切线方程是________.
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