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58th derivative of (1+x^3)^30 at x=0

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    find the 58th derivative of (1+x^3)^(30) at zero (The Answer is zero)

    2. Relevant equations

    The binomial series expansion
    the maclaurin series (f^(n)(0)/n!)x^n

    3. The attempt at a solution
    So I expanded the given equation into the binomial series and at at the 19th term, I got 30(30-1)...(30-19+1)/57! * x^(57). But I don't get how that gets me any closer to the answer of zero.
     
  2. jcsd
  3. Nov 7, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    First just figure out what powers of x could have a nonzero 58th derivative at zero. Try a few.
     
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