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

Question

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

Answer 1

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