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Partial derivative of integral with variable limit

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] G(\theta, k ) = \int^{\theta}_0 g(x,k) dx [/tex]

    [tex] \frac{\partial G}{\partial \theta} = ? [/tex]

    [tex] \frac{\partial G}{\partial k} = ? [/tex]

    3. The attempt at a solution

    If I say that [tex] \int g(x,k) dx = H(x,k) [/tex]

    [tex] \int^{\theta}_0 g(x,k) dx = H(\theta,k) - H(0,k) [/tex]


    Then is [tex]\frac{\partial G}{\partial \theta} = \frac{\partial H(\theta,k)}{\partial \theta}=g(\theta,k)[/tex]
    ?
     
  2. jcsd
  3. Jun 12, 2010 #2

    vela

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    Yup.
     
  4. Jun 13, 2010 #3

    HallsofIvy

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    That is "correct" but your derivation is more complicated than necessary! By the fundamental theorem of calculus, the derivative with respect to [itex]\theta[/itex] is just [itex]g(\theta, k)[/itex].

    Lagrange's formula is an extension of the fundamental theorem of calculus:
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t) dt= f(x, \beta(x))}\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}dt[/tex].

    Since [itex]\theta[/itex] only appears in the upper limit of the integral, the derivative with respect to [itex]\theta[/itex] is just
    [tex]f(\theta, k)\frac{d\theta}{d\theta}= f(\theta, k)[/tex]

    and the derivative with respect to k is just
    [tex]\int_0^\theta \frac{\partial f}{\partial k} dx[/tex]
     
  5. Jun 13, 2010 #4
    That's clearer now, thank you.
     
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