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Derivative of an integral

  1. Feb 11, 2012 #1
    Hi all,

    This is my question. Suppose that I have a continuum of people on the measure [0,1]. Then, I want to aggregate their spending as follows,

    [itex]\int^{1}_{0}(s^{\alpha}(i))di[/itex]

    where i is any person in [0,1]. Suppose that I want to compute the derivative of the previous expression if [itex]s^{\alpha }[/itex] goes up for only one specific i[itex]\in[0,1][/itex]. That is, I want,

    [itex]\frac{d(\int^{1}_{0}(s^{\alpha}(i))di)}{d(s(i))}[/itex]

    How can I compute that derivative...? First, I thought that it would be zero since the contribution to the integral is infinitesimal, but I am not sure about that...

    Thank you!
     
    Last edited: Feb 11, 2012
  2. jcsd
  3. Feb 11, 2012 #2

    HallsofIvy

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    Is [itex]s^{\alpha}[/itex] s to the [itex]\alpha[/itex] power? If so Leibniz's rule:
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x,\beta(x))- \frac{d\alpha}{dx}f(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x}dt[/tex]
    Works: the derivative is
    [tex]\alpha \int_0^1 s^{\alpha- 1}(i)di[/tex]
     
  4. Feb 11, 2012 #3
    Sorry, I did not write the problem properly. There is one specific J[itex]\in(0,1)[/itex], and I want
    [itex]\frac{d(\int^{1}_{0}(s(i)^{\alpha})di)}{d(s(J))}[/itex]. That is, the derivative with respect to that specific s(J). And yes, it is to the power of [itex]\alpha[/itex].
     
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