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Find the 3rd derivative using MacLaurin Series

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = e^(x^2) * sin(x)
    Find the value of the 3rd derivative at x = 0.

    2. Relevant equations

    e^x = 1 + x + x^2/2! + ... + x^n/n!

    sin(x) = 1 + x^3/3! - x^5/5! + ... + x^(2n+1)/(2n+1)! * (-1)^(n-1)


    3. The attempt at a solution

    I know I should plug in the two series into f(x). But what is e^(x^2)? Would I have to basically square the power series of e^x?

    So let's just make n = 3, then
    f(x) = (1 + x + x^2/2! + x^3/3!)^2 * (1 + x^3/3! - x^5/5! + x^7/7!)

    Then the expression becomes extremely complex for me. Even if I manage to expand the whole thing, how would I use it to find the third derivative?
     
  2. jcsd
  3. Nov 11, 2009 #2

    Mark44

    Staff: Mentor

    For your Maclaurin series for e^x, just replace x by x^2 in each term on the right. That will be your Maclaurin series for e^(x^2).

    Your series for sin(x) is wrong: sin(x) = x - x^3/3! + x^5/5! -+ ...

    Your series for f(x) is going to be the product of your two series. You don't need all that many terms, since you just need the coefficient of the 3rd degree term.
     
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