Question

求函数导数f(x)=ln ctg x/2 在点x= -π/6 .

Answer 1

f(x)=lnctg(x/2);

f'(x)=[1/ctg(x/2]*[ctg(x/2)]'

=[1/ctg(x/2)]*[-csc^2(x/2)]*(x/2)'

=-[csc^2(x/2)/[2ctg(x/2)]

=-1/[2sin(x/2)cos(x/2)]

=-1/sinx.

f'(-π/6)

=-1/sin(-π/6)

=1/sin(π/6)

=2.