Question

求导求导求导

Answer 1

解：g'=((2x-2)(x-lnx)-(x^2-2x)(x-1/x))/(x-lnx)^2
=((2x-2)(x-lnx)-(x^2-2x-(x-2))/(x-lnx)^2
=(x-1)(x-2lnx+2)/(x-lnx)^2.
答：倒数为(x-1)(x-2lnx+2)/(x-lnx)^2。

Answer 2

g(x)＝(x^2-2x)/(x-lnx)

g'(x)＝[2(x-1)(x-lnx)-(x^2-2x)(1-1/x)]/(x-lnx)^2
＝[2(x-1)(x-lnx)-(x-2)(x-1)]/(x-lnx)^2
＝(x-1)(x+2-2lnx)/(x-lnx)^2