f(x) = (3x+1)^3
f'(x) = 3(3x+1)^2(3)
f'(x) = 9(3x+1)^2
f"(x) = 9[(2)(3x+1)(3)] + [9(3x+1)^2]
f"(x) = 54(3x+1) + 9(3x+1)^2
f"(x) = 9(3x+1)[6 + (3x+1)^2]
f"(x) = 9(3x+1)(7+3x)
Am I correct?
You are right until you decided to take the second derivative:
f(x) = (3x+1)^2
f'(x) = 3(3x+1)^2*(3)
Now in order to get f''(x) you must use the product rule (not the triple product rule as constants hold over) thus:
f''(x) = (3)*(2(3x+1))*(3)*(3)
You do not have to but I choose to simplify so
f''(x)= 54(3x+1)