设y=lntan(x/2)，求y'

来源：百度知道 编辑：UC知道 时间：2024/06/11 01:28:40
设y=lntan(x/2)，求y'

答案居然是cscx
我怎么也算不出这个答案来

y'={1/[tan(x/2)]}*[tan(x/2)]'
={1/[tan(x/2)]}*[1/cos^2(x/2)]*(x/2)'
={1/[tan(x/2)]}*[1/cos^2(x/2)]*(1/2)
=[cos(x/2)/sin(x/2)]*[1/cos^2(x/2)]*(1/2)
=1/2sin(x/2)cos(x/2)
=1/sinx
=cscx

一共有3层，它是个复合函数，要从外层往里层一步步求

y'=1/tan(x/2)*(tan(x/2))'
=cos(x/2)/sin(x/2)*1/cos(x/2)^2*1/2
=1/[sin(x/2)*cos(x/2)*2]
=1/sinx
=cscx