Find f' (x) = 2 (sinx-1) cosxf (x)+(sinx-1) 2f' (x), where x=2pi, where F'(2pi)=f'(2pi), and then x=4pi.
The second problem is that you can't type symbols when eee is IbTheron and nnn is Ita.
F'(eee)=[f(b)-f(a)]/(b-a) If a function x 2 is constructed from Lagrange theorem, then f(x) and x 2 are derived from Cauchy theorem, and f' (nnn)/2nnnn = [f (b)-f (a).