Question 428180
I'll get you started:


If f(x) = {{{x^3 = 3x^2 + x + 1}}}, then f'(x) = {{{3x^2 + 1}}} and f''(x) = {{{6x}}} (simply applying the power rule).

If f(x) = {{{sin(x)+1}}} then f'(x) = {{{cos(x)}}} and f''(x) = {{{-sin(x)}}}.

#3 is a little harder to differentiate because we need to apply the product rule twice. If f(x) = {{{(x^2)(e^(-4x))}}} then f'(x) = {{{2x(e^(-4x)) - 4(x^2)(e^(-4x)) = e^(-4x)(2x - 4x^2)}}}. Differentiating again, f''(x) = {{{-4e^(-4x)(2x - 4x^2) + e^(-4x)(2 - 8x) = e^(-4x)(-8x + 16x^2 + (2 - 8x)) = e^(-4x)(16x^2 - 16x + 2)}}}. I'll let you handle the 24 parts of the problem.