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Differential of f(x+dx)

  1. Dec 1, 2011 #1
    Hi all,

    I'm a bit stuck on what should probably be fairly simple, but I'm looking for a general way to do

    [itex]\partial_{x}f(x+\delta x)[/itex]

    Any help would much appreciated.
  2. jcsd
  3. Dec 1, 2011 #2


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    The notation is a little off beat. I've never seen partial differentials, only partial derivatives.
  4. Dec 1, 2011 #3
    I'm not quite sure what is meant either. Is δ a number or a function. Is f a function of more than one variable (if not why the partial derivative notation)?
  5. Dec 2, 2011 #4
    Sorry, just some old habits. What I mean is this: A function f(x) is perturbed by δx so f(x+δx). What is the differential of this:

    [itex]\frac{d}{dx}f(x+\delta x)[/itex]

    which is what I think it should actually look like.
  6. Dec 2, 2011 #5

    Stephen Tashi

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    If you mean [itex]\delta x [/itex] to be a finite constant then this is like asking "What is the differential of f(x+5)?" Is it that sort of question?

    Or are you doing some sort of reasoning involving "infinitesimals"? If so, it would be better to give a complete context for the situation.
  7. Dec 5, 2011 #6
    Ok, if I'm going to describe the full problem this should go in the cosmology section.

    The goal is to rewrite the Klein-Gordon equation in terms of a perturbation to the scalar field. So, starting from

    [itex]\frac{d^{2}\phi}{dt^{2}} + 3H\frac{d\phi}{dt} + \frac{dV}{d\phi}[/itex]=0

    where [itex] \phi=\phi(x,t) [/itex] and [itex] V=V(\phi) [/itex]

    and using

    [itex] \phi(x,t)=\phi(t)+\delta\phi(x,t) [/itex]

    I have gotten as far as

    [itex] \frac{d^{2}\phi}{dt^{2}} + \frac{d^{2}\delta\phi}{dt^{2}} +3H\frac{d\phi}{dt} + 3H\frac{d\delta\phi}{dt} + \frac{dV(\phi + \delta\phi)}{d\phi} = 0[/itex]

    Now what I am trying to get is

    [itex] \frac{d^{2}\delta\phi}{dt^{2}} + 3H\frac{d\delta\phi}{dt} +\frac{d^{2}V}{d\phi^{2}}\delta\phi =0[/itex]

    So you see what I meant with my original post. I figured If I could evaluate the last term I'd get the correct answer but I can't remember how to do it.
  8. Dec 5, 2011 #7
    you should now subtract the original unperturbed equation and apply the differentiation rule: df/dx = f(x+dx) - f(x) which immediately leads to the result.
  9. Dec 6, 2011 #8
    It worked thanks allot. I knew it would be something easy.
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