Question

f'(x) =(3cos 4x - 28x² +6,7x - 40,8)'

Answer 1

$f'(x)=(3cos4x - 28x^2 +6,7x - 40,8)' \\ (3cos4x)'=-12sin4x \\ (28x^2)'=56x \\ (6,7x)=6,7 \\ (40,8)'=0 \\ f'(x)=-12sin4x-56x+6,7$