Keep the formula for the Maclaurin series in mind: f(x) = sum (n=0 to infinity) of [ f(n)(0) / n! ] * x^n.
f(n) is the nth derivative of f , n! is n factorial and f(n)(0) is the nth derivative of f evaluated at 0.
Then the Cn's are the f(n)(0) / n! in the above formula. with f(0)(0) is the zeroth derivative or just the function itself. since x^0 is 1 by definition,
C0 is f evaluated at zero or f(0)(0) = 0 since there is a leading x term.
C1 - you have to take the first d/dx of the function = [-6x*e^(-3x) ] * [3x-2] and evaluate it at x = 0 then divide by n!. You get 0
C2 - second derivative gives : 6e^(-3x) * (9x^2 -12x +2) / 2! = 6. So the second term in the series is 6x^2 from f(2)(0)/2! * x^2
C3 - third derivative gives : -54e^(-3x) * (3x^2 -6x +2). evaluate at x=0 gives, -108. which if you divide by 3! is -18. so C3 is 18, and the third term is -18x^3
Similarly, C4 = 27, C5 =-27
The thing with this problem is the massive and time consuming derivative taking. Just write down the function, then the first derivative, then the second etc being extremely careful to use the product rule well. Simplify each time you do so and practice doing this. I can't imagine that a prof would give this in an exam since it would just test product rule speed more than actual understanding. If it does come up, write down the formula for the Maclaurin series, explain the C's come from the leading coefficients of each term and that you understand that there are 5 derivatives to take.
Then do your best to get them done, but if time is important, do another problem and finish the last 2 derivatives later.
Marc D.
That is way easier. Do that!11/21/20