the answer is #f'(pi/12)# where #f'(x)# is the derivative of #f(x)#
first simplify the #-(cos(x-pi))^2# part:
#=-(-cos(x))^2# proof
#=-(cosx)^2#
so #f(x)=secx-(cosx)^2#
#f'(x)=d/dx(secx)-d/dx((cosx)^2)#
#f'(x)=secxtanx-(2cosx*d/dx(cosx))# (secx derivative and chain rule)
#f'(x)=secxtanx-2cosx(-sinx)# (cosx derivative)
#f'(x)=secxtanx+2sinxcosx#
#f'(x)=secxtanx+sin(2x)# (double angle formula for sin(2x))
now evaluate at #x=pi/12#
#f'(pi/12)=sec(pi/12)tan(pi/12)+sin(2*(pi/12))#
use a calculator, memorize these, or use this table to evaluate:
#f'(pi/12)=(sqrt(6)-sqrt(2))*(2-sqrt(3))+sin(pi/6)#
#f'(pi/12)=2sqrt(6)-2sqrt(2)-sqrt(18)+sqrt(6)+1/2#
#f'(pi/12)=3sqrt(6)-2sqrt(2)-3sqrt(2)+1/2#
#f'(pi/12)=3sqrt(6)-5sqrt(2)+1/2#
