令y=lnu,u=tanv.v=x/2
所以y'=(lnu)'(tanv)'(x/2)'
=(1/u)(sinv/cosv)'(1/2)
=(1/2)[1/(tanx/2)]{[(sinv)'cosv-sinv(cosv)']/(cosv)^2}
=(1/2)[1/(tanx/2)]{[(cosv)^2+(sinv)^2]/(cosv)^2}
=(1/2)[1/(tanx/2)][1/(cosx/2)^2]
=(1/2)[(cosx/2)/(sinx/2)][1/(cosx/2)^2]
=(1/2){1/[(sinx/2)(cosx/2)]}
=1/[2(sinx/2)(cosx/2)]
=1/sinx