f(1)
=1^2-2/1+ln(1/1)
=1-2+ln1
=-1+0
=-1应该f(1)’是多少谢谢f(x)
=(1/x)^2-2x+lnx
=1/x^2-2x+lnx
f(x)'
=-2x^(-3)-2+1/x
所以
f(1)'
=-2-2+1
=-4+1
=-3