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

  • Thread starter graupner1000
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In summary, the goal is to rewrite the Klein-Gordon equation in terms of a perturbation to the scalar field. This is done by taking the derivative of the equation with respect to time and solving for the derivative of the perturbation.
  • #1
graupner1000
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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

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

Any help would much appreciated.
 
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  • #2
The notation is a little off beat. I've never seen partial differentials, only partial derivatives.
 
  • #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)?
 
  • #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.
 
  • #5
graupner1000 said:
A function f(x) is perturbed by δx

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.
 
  • #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.
 
  • #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.
 
  • #8
It worked thanks allot. I knew it would be something easy.
 

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

What is the definition of differential of f(x+dx)?

The differential of f(x+dx) is a mathematical concept used in calculus to represent the change in a function f(x) as its input x changes by a small amount dx. It is also known as the derivative.

How is the differential of f(x+dx) calculated?

The differential of f(x+dx) is calculated using the limit definition of a derivative. This involves finding the slope of a tangent line to the function f(x) at a given point x and using that slope to approximate the change in the function as x changes by a small amount dx.

Can the differential of f(x+dx) be negative?

Yes, the differential of f(x+dx) can be negative. This indicates that the function f(x) is decreasing at that point and the change in the function is in the negative direction.

How is the differential of f(x+dx) used in real-world applications?

The differential of f(x+dx) is used in various fields of science and engineering, such as physics, economics, and engineering, to model and analyze real-world phenomena. It is particularly useful in predicting the behavior of systems that change continuously over time.

Are there any limitations to using the differential of f(x+dx)?

Yes, there are limitations to using the differential of f(x+dx). It is only an approximation of the actual change in the function and becomes less accurate as the value of dx increases. Additionally, it cannot be used for functions that have sharp turns or corners, as the derivative does not exist at these points.

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