Question

f'(x) =(3cos 4x - 28x² +6,7x - 40,8)'

Answer 1

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