Question

求导数，怎么做啊

设f'(x)=lnx, 令y=f[(2x-1)/(x+1)], 求dy/dx

Answer 1

设y=f(u)

dy/dx =dy/du*du/dx=d(lnu)/du *du/dx=1/u *du/dx=1/u *[2(x+1)-(2x-1)]/(x+1)^2
=3/[(2x-1)(x+1)]

Answer 2

复合函数求导法。
dy/dx =ln[(2x-1)/(x+1)][(2x-1)/(x+1)]'
=[3/(x+1)^2]ln[(2x-1)/(x+1)]

Answer 3

dy/dx
= d(f[(2x-1)/(x+1)])/dx
={f'[(2x-1)/(x+1)]}{(2x-1)-2(x+1)}/(x+1)^2
=-3{f'[(2x-1)/(x+1)]}/(x+1)^2

f'(x) = lnx
f'[(2x-1)/(x+1)] = ln{ (2x-1)/(x+1) }

dy/dx
=-3ln{ (2x-1)/(x+1) }/(x+1)^2