题目内容
曲线y=在点(1,2)处的切线为________.
曲线y=在点(1,1)处的切线方程为
A.x-y-2=0
B.x+y-2=0
C.x+4y-5=0
D.x-4y-5=0
设曲线y=在点(3,2)处的切线与直线ax+y+1=0垂直,则a=
2
-2
-
过点(0,1)且与曲线y=在点(3,2)处的切线垂直的直线的方程为
A.x+2y-2=0
B.2x+y-1=0
C.2x-y+1=0
D.x-2y+2=0