y=[x(x+1)/(x+2)e^x]^1/3怎么求其微分.

等式两端取自然对数,有：
lny=(1/3)[lnx+ln(x+1)-ln(x+2)-x]
等式两端同时对x求导,有：
1/y*y导=(1/3)[1/x+1/(x+1)-1/(x+2)-1]
所以：y导=[x(x+1)/(x+2)e^x]^(1/3)*(1/3)[1/x+1/(x+1)-1/(x+2)-1]
所以：dy={[x(x+1)/(x+2)e^x]^(1/3)*(1/3)[1/x+1/(x+1)-1/(x+2)-1]}dx