设f(x)=ln(tanx)
则
f'(x)=1/tanx *(tanx)'
=1/tanx *1/cos^2 x
=cosx/sinx *1/cos^2 x
=1/(sinxcosx)
=2/sin2x
f''(x) =-2 *2*cos2x/sin^2 2x
=-4cos2x/sin^2 2x
设f(x)=ln(tanx)
则
f'(x)=1/tanx *(tanx)'
=1/tanx *1/cos^2 x
=cosx/sinx *1/cos^2 x
=1/(sinxcosx)
=2/sin2x
f''(x) =-2 *2*cos2x/sin^2 2x
=-4cos2x/sin^2 2x