Question

用对数求导法求导：y=[x*(x^2+1)/(x-1)^2]^1/3

Answer 1

用对数求导，
等式两边同时取对数，得到
lny=lnx
-
ln(1+x^2)
等式两边同时对x求导，得到
y'/y=1/x
-2x/(1+x^2)
所以
y'=y*[1/x
-2x/(1+x^2)]
再带入y=x/(1+x^2)
所以
y'=x/(1+x^2)
*
[1/x
-2x/(1+x^2)]
=1/(1+x^2)
-2x^2/(1+x^2)^2
=(1-x^2)/(1+x^2)^2

Answer 2

等式两边取对数有：lny=1/3ln[x*(x^2+1)/(x-1)^2]
化简得3lny=lnx+ln(x^2+1)-2ln|x-1|
两边求导3y'/y=1/x+2x/(x^2+1)-2/(x-1)
y'=1/3*y[1/x+2x/(x^2+1)-2/(x-1)]
=1/3*[1/x+2x/(x^2+1)-2/(x-1)]*[x*(x^2+1)/(x-1)^2]^1/3