f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]
=(1+cos2x)/(3-cos2x)
=-1+4/(3-cos2x)
f'(x)=-4/(2-cos2x)^2*(2-cos2x)'
=8sin2x/(2-cos2x)^2
所以f'(π/4)=2