若∏/4＜x＜∏/2,则函数y=tan2x(tanx)^3的最大值是多少

y=2(tanx)^4/[1-(tanx)^2]
令t=(tanx)^2
因为tanx>1
t>1
y=2t^2/(1-t)
=2(t^2-1+1)/(1-t)
=2[(t^2-1)/(1-t)+1/(1-t)]
=2[-t-1-1/(t-1)]
=-2[t+1+1/(t-1)]
=-2[2+(t-1)+1/(t-1)]
因为t-1>0
所以(t-1)+1/(t-1)》2
y《-2(2+2)
y《-8