How do I calculate the differential of f(x+dx)?

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Hi all,

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

\partial_{x}f(x+\delta x)

Any help would much appreciated.
 
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The notation is a little off beat. I've never seen partial differentials, only partial derivatives.
 
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)?
 
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:

\frac{d}{dx}f(x+\delta x)

which is what I think it should actually look like.
 
graupner1000 said:
A function f(x) is perturbed by δx

If you mean \delta x 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.
 
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

\frac{d^{2}\phi}{dt^{2}} + 3H\frac{d\phi}{dt} + \frac{dV}{d\phi}=0

where \phi=\phi(x,t) and V=V(\phi)

and using

\phi(x,t)=\phi(t)+\delta\phi(x,t)

I have gotten as far as

\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

Now what I am trying to get is

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

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.
 
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.
 
It worked thanks allot. I knew it would be something easy.
 
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