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If f(x) = sin(x^3) find f^(15)(0)

  1. Aug 9, 2008 #1
    Short and sweet, If f(x) = sin(x^3) find f^(15)(0) by which I mean find the 15th derivative of f. Do I have to write out all the derivatives and look for a pattern somewhere or is there an easier way?
     
  2. jcsd
  3. Aug 9, 2008 #2

    gel

    User Avatar

    Do you know the power series expansion for sin? If so, you can read off the derivatives from the Taylor series expansion
    [tex]
    f(x) = \sum_{n=0}^\infty\frac{x^n}{n!}f^{(n)}(0).
    [/tex]
     
  4. Aug 9, 2008 #3
    Ok, so from the maclaurin series with x^3 substituted I get f(x) = x^3 - (x^9/ 3!) + (x^15/ 5!) and just from this I can see the 15th derivative of the function is going to be 15!/ 5!. Is that all correctÉ
     
  5. Aug 9, 2008 #4

    gel

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    yes!
     
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