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Derivatives wrt functions

  1. Aug 17, 2010 #1

    Char. Limit

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    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]
     
  2. jcsd
  3. Aug 17, 2010 #2
    Last edited by a moderator: Apr 25, 2017
  4. Aug 17, 2010 #3

    alxm

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    Sounds like you're looking for http://en.wikipedia.org/wiki/Functional_derivative" [Broken].
     
    Last edited by a moderator: May 4, 2017
  5. Aug 17, 2010 #4
    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.
     
  6. Aug 17, 2010 #5
    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.
     
    Last edited by a moderator: May 4, 2017
  7. Aug 18, 2010 #6

    alxm

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    Ah yes, silly me.. Guess I've been reading too many DFT papers lately.
     
  8. Aug 18, 2010 #7

    Pengwuino

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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.
     
  9. Aug 18, 2010 #8
    It's basically the method of substitution "on the go".
     
  10. Aug 18, 2010 #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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