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Nth Derivative of a Function

  1. Aug 26, 2013 #1

    FeDeX_LaTeX

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    1. The problem statement, all variables and given/known data
    Let ##f(x) = \frac{\sin x}{b + \cos(ax)}##. Show that the nth derivative ##f^{(n)}(0) = 0## if n is an even integer.


    2. Relevant equations
    Leibniz's generalised product rule:

    ##(f \cdot g)^{(n)} = \sum_{k = 0}^{n} \binom{n}{k} f^{(k)}g^{(n-k)}##


    3. The attempt at a solution
    I'm letting ##f(x) = \sin x## and ##g(x) = \frac{1}{b + \cos(ax)}## then applying Leibniz's rule. Clearly, the terms of the series k = 0, k = 2, ... (every even k) are all 0 when x = 0, since they all contain an even derivative of sin (which gives us sin again). But what do I do about the derivatives of g(x)? Is this the right approach?
     
  2. jcsd
  3. Aug 26, 2013 #2
    Odd functions of x have only odd powers of x in their Taylor-McLaurin series, so...
     
  4. Aug 27, 2013 #3

    FeDeX_LaTeX

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    So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?
     
  5. Aug 27, 2013 #4

    Ray Vickson

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    There is a much, much easier way. Just answer the following three questions.
    1. The given function is (a) odd; (b) even; (c) neither.
    2. The derivative of an even function is (a) even; (b) odd; (c) neither.
    3. The derivative of an odd function is (a) even; (b) odd; (c) neither,
     
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