2.
a) y'=x'·cosx+x·(cosx)' = 1·cosx+x·(-sinx) = cosx - x·sinx
b) y' = [(tgx)'·x -tgx·x']/x² = [1/cos²x·x -tgx·1]/x² = 1/(x³cos²x) - tgx/x²
c) y' = 4(3x+5)³ ·(3x+5)' = 12(3x+5)³
3.
f'(x) = 2cosx +6x -2π
f'(π/3) =2cosπ/3 +3·(π/3) -2π = 2·1/2 +π -2π =1 - π