曲线y=lnx上与直线x+y=1垂直的切线方程为
解题思路:由题目条件可知道,切线斜率为1,也就是切点处导数为1,这样能确定切点坐标,再利用点斜式即可.由y′=(lnx)′=
1
x=1,得x=1,可见切点为(1,0),于是所求的切线方程为y-0=1•(x-1),即 y=x-1.
故答案为:y=x-1
,4,
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