Find the coefficient of the squared term in the simplified form for the second derivative, f "(x) for f(x) = (x3 + 3x2 + 3)(3x3 − 6x2 − 8x + 1)

Hmm

ok so some rules
(f(x)+g(x))'=f'(x)=g'(x)
and
f'(c)=0 where c is a constant

(g(x)f(x))'=f'(x)g(x)+g'(x)f(x)
and


so
f'(x)=(x^3+3x^2+3)(3x^3-6x^2-8x+1)=
f'(3x^6+3x^5-26x^4-14x^3-15x^2-24x+3)=
f'(3x^6)+f'(3x^5)-f'(26x^4)-f'(14x^3)-f'(15x^2)-f'(24x)+f'(3)=
18x^5+15x^4-104x^3-42x^2-30x-24
notice the powers went down
focus on the one that will have x^2 so it has x^3 righ tnow and take the derivitive of it

f'(-104x^3) =-312x^2

the coefieint is -312