Derivatives wrt functions

  • #1
Char. Limit
Gold Member
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Main Question or Discussion Point

Is it possible to take a derivative with respect to a function, rather than just a variable? I'll give a simple example of how I imagine such a thing would work to try to explain...

[tex]\frac{d}{d(sin(x))}\left(sin^2(x)\right) = 2 sin(x)[/tex]

Can you take a derivative this way?

Also, can you write an equivalent integral as such?

[tex]\int 2 sin(x) d(sin(x)) = sin^2(x) + C[/tex]
 

Answers and Replies

  • #3
alxm
Science Advisor
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Sounds like you're looking for http://en.wikipedia.org/wiki/Functional_derivative" [Broken].
 
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  • #4
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To differentiate f(x) w.r.t g(x), just do the following:

[tex]\frac{d f(x)}{d g(x)}[/tex]

[tex]=\frac{d f(x)}{d x}\frac{d x}{d g(x)}[/tex]

[tex]=\frac{d f(x)}{d x}\left(\frac{d g(x)}{d x}\right)^{-1}[/tex]
Char. Limit, for the integration, what you have written is the standard way I learnt to do integration. The general form is:

[tex] \int g(x) f'(x)d x=\int g(x)d f(x)[/tex]

For example, by using the property of differential, [tex] d x^2/2 = x d x [/tex]

[tex]\int e^{x^2}x \text{dx}[/tex]

[tex]=\frac{1}{2}\int e^{x^2}\text{dx}^2[/tex]

[tex]=\frac{1}{2}e^{x^2}+C[/tex]

This method of doing integration is much better than using substitution in many situations. Since if the integral is complicated, you don't have to do substitution repeatedly.
 
  • #5
2,967
5
Sounds like you're looking for http://en.wikipedia.org/wiki/Functional_derivative" [Broken].
no, it is just parametric ordinary differentiation.

[tex]
w = f(u) = u^{2}, u(x) = \sin{(x)} \Rightarrow \frac{dw}{du} = f'(u) = 2 u = 2 \sin{(x)}
[/tex]

Functional derivatives are derivatives of functionals with respect to functions. [itex]\sin^{2}{(x)}[/itex] is not a functional.
 
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  • #6
alxm
Science Advisor
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Functional derivatives are derivatives of functionals with respect to functions. [itex]\sin^{2}{(x)}[/itex] is not a functional.
Ah yes, silly me.. Guess I've been reading too many DFT papers lately.
 
  • #7
Pengwuino
Gold Member
4,989
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You'll even see things such as integrals over volumes, [tex]d^3x[/tex], done in terms of [tex]d(cos(\theta))[/tex] instead of just [tex]d\theta[/tex]. So instead of [tex]r^2sin(\theta)drd\theta d\phi[/tex], you can change your limits of integration to be [tex]r^2drd(cos(\theta))d\phi[/tex]. My first encounter, and I bet a lot of physics majors first encounter, is integrating Legendre Polynomials that are dependent on [tex]cos(\theta)[/tex] so it is only natural.
 
  • #8
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5
It's basically the method of substitution "on the go".
 
  • #9
HallsofIvy
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It's really just the chain rule. The "derivative of f(x) with respect to g(x)" is
[tex]\frac{df}{dg}= \frac{df}{dx}\frac{dx}{df}= \frac{\frac{df}{ex}}{\frac{dg}{dx}}[/tex].
It is simply "the rate of change of f compared to that of g" or "the rate of change of f divided by the rate of change of g".

And, as others have said, integration in that form is just "substitution" which is, itself, the "chain rule in reverse".
 

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