1.设y^(n-1)=e^x+cos2x,
∴y^(n)=e^x-2sin2x,
y^(n+1) =e^x+4cos2x.
2.lim < x→π/2> (x-π/2)cot2x
=lim < x→π/2> (x-π/2)/tan2x
=lim < x→π/2>1/[2(sec2x)^2]
=1/2.第一题好像不对~~~第二题对了。。。y^(n-1)=e^x+cos2x，两边对x求导得y^(n)=e^x-2sin2x,再两边对x求导得y^(n+1) =e^x+4cos2x.最后应该是减号吧？。。。您说的对：y^(n+1) =e^x-4cos2x.