对t求导：x=a(lntant/2+cost)答案是x’=acos^2t/sint是x=a(lntan(t/2)+cost)

Question

对t求导：x=a(lntant/2+cost)
答案是x’=acos^2t/sint
是x=a(lntan(t/2)+cost)

Answer 1

先不写a
最后乘
cost'=-sint
[lntan(t/2)]'
=[1/tan(t/2)]*[tan(t/2)]'
=[1/tan(t/2)]*sec^2(t/2)*(t/2)'
=(1/2)sec^2(t/2)/tan(t/2)
=(1/2)[1/cos^2(t/2)]/[sin(t/2)/cos(t/2)]
=1/[2sin(t/2)cos(t/2)]
倍角公式
=1/sint
相加
x'=a[1/sint - sint]
通分
=a[1-sin^2 t]/sint
=acos^2 t / sint