Nth Derivative of an Even Function

FeDeX_LaTeX
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Homework Statement


Let ##f(x) = \frac{\sin x}{b + \cos(ax)}##. Show that the nth derivative ##f^{(n)}(0) = 0## if n is an even integer.


Homework Equations


Leibniz's generalised product rule:

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


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?
 
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Odd functions of x have only odd powers of x in their Taylor-McLaurin series, so...
 
So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?
 
FeDeX_LaTeX said:
So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?

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,
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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