Question

f(x)=1/(x-lnx)，如何求得x>1时f(x)的导函数<0？

f(x)=1/(x-lnx)，如何求得x>1时f(x)的导函数<0？

Answer 1

f(x)=1/(x-lnx)
f'(x)=-(x-lnx)'/(x-lnx)²
=-(1- 1/x)/(x-lnx)²
=(1-x)/[x(x-lnx)²]
f'(x)=0时有极值
(x-lnx)²>0
1-x=0，x=1
f'(x)>0时，f(x)单调↑
1-x>0→0<x<1
f'(x)<0，x>1，f(x)单调↓

Answer 2

f(x)=1/(x-lnx)
f'(x)=-(x-lnx)'/(x-lnx)²
=-(1- 1/x)/(x-lnx)²
=(1-x)/[x(x-lnx)²]
(x-lnx)²>0
x>1，x>0，1-x<0
(1-x)/[x(x-lnx)²]<0，f'(x)<0
即x>1时，f(x)的导函数<0

Answer 3

求导，分母恒大于0，分子为1/x-1，x大于1时导数小于0

Answer 4

先求导，令导数＜0你就知道了

Answer 5

主题曲的下载