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Differentiation with different variables

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to take the derivative of the following integral

    [itex]\frac{d}{d V} \int_0^t{V(\tau)}d\tau[/itex]

    2. Relevant equations

    FTC will probably be a part of it.

    3. The attempt at a solution

    I always get confused when I'm taking the derivative of an integral. I know the answer is not simply [itex] V(t) [/itex] though since the integrating variable is not the same as the variable I am trying to take the derivative with respect to.
     
  2. jcsd
  3. Mar 4, 2013 #2
    I found the expression [itex]\frac{d}{d V} \int_0^t{V(\tau)}d\tau[/itex] a little surprising. I'm not sure how to interpret the [itex]\frac{d}{d V}[/itex]. I would have expected to see [itex]\frac{d}{d t} \int_0^t{V(\tau)}d\tau[/itex] instead. In that case it is exactly the fundamental theorem of calculus and the solution is ##V(t)##.

    Note that I would expect the solution to be a function of ##t## because the expression itself is a function of ##t##. ##τ## is a dummy variable used to compute the definite integral - the expression itself is not a function of ##τ##.
     
  4. Mar 4, 2013 #3
    Yes, I agree it is weird, but it is part of a calculus of variations problem, and is not a typo or anything.
     
  5. Mar 4, 2013 #4
    Thinking more about the problem, if it is legal to bring the derivative inside the integral, then the answer would just be [itex]t[/itex]. This seems odd in the context of the larger problem, though.

    Can anybody comment on if this is correct or not. Thanks.
     
  6. Mar 4, 2013 #5
    That is not legal. The only thing you can move in and out of integrals is contants.
     
  7. Mar 4, 2013 #6

    CAF123

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    Gold Member

    You could perhaps start with the chain rule.
     
  8. Mar 4, 2013 #7

    Fredrik

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    You may need to provide more information. What does d/dV even mean here? It can't be a derivative, since the thing to the right of it isn't the value at V of some differentiable function. Is it a functional derivative? Does the course you're taking cover the definition of "functional derivative"? Does the book prove a theorem for functional derivatives that is similar to the chain rule?
     
  9. Mar 4, 2013 #8

    Ray Vickson

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    In Calculus of variations and Optimal Control we talk about "functional derivatives", which are like directional derivatives in function space. So, given a functional
    [tex] F(f) = \int_a^b L(f(t)) \, dt,[/tex]
    the directional derivative of F(f) in the direction h (also called the Gateaux derivative) is
    [tex] DF(f;h) \equiv \frac{d}{d\epsilon} \int_a^b F(f(t) + \epsilon h(t)) \, dt\\
    = \int_a^b F^{\prime}(f(t))\cdot h(t) \, dt.[/tex]
    In this notation, your functional ##F(V) = \int_0^t V(\tau) \, d\tau## would have directional derivative
    [tex] DF(V;h) = \int_0^t h(\tau) \, d\tau.[/tex]

    I don't know if that is what you want, but that is what is done in all treatments of Calculus of Variations that I have seen.
     
  10. Mar 4, 2013 #9
    [itex]\frac{d}{dV}[/itex] is the derivative with respect to the function [itex]V(t)[/itex].

    This is a calculus of variations problem, and the functional is

    [itex]\int_0^{t_1}{\mathcal{M}}dt[/itex],

    [itex] \mathcal{M} = \int_0^t{V(\tau)}d\tau + \text{other stuff...} [/itex]

    What I'm trying to do, is find the equation for


    [itex] \frac{d\mathcal{M}}{dV} = \frac{d}{dt} \frac{d\mathcal{M}}{d\dot{V}}[/itex]

    which is where the derivative comes from in the first place.

    Hope this helps.

    Thanks.
     
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