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Finding Partial Derivative of an Integral

  1. Mar 5, 2008 #1
    Hey everyone. Im new here and i have a problem with some partials. We're studying partial derivatives in calculus III. I understand and all, but we haven't covered how to take a partial derivative of an integral. This problem showed up in my practice problems before our exam tomorrow.

    The problem is as follows:

    [tex]\frac{\partial}{\partial}x[/tex][tex]\int[/tex]cos(t[tex]^{3}[/tex])dt

    If you cant follow that.. then it says we need the first partials (x and y) of the integral of cos(t^3)dt.. (lower limit=y, and upper limit=x)

    Any insight on how to start this problem?? Thanks for any help!!


    -Kev
     
  2. jcsd
  3. Mar 5, 2008 #2

    HallsofIvy

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    I couldn't follow it because you didn't put in the limits of integration, and that is crucial!

    You should know from single variable calculus, the "Fundamental Theorem of Calculus":
    [tex]\frac{d}{dt}\int_a^x f(t)dt= f(x)[/tex]
    where a is any constant.
    From that it should be easy to find the partial derivative with respect to x.

    To find the derivative with respect to y, remember that
    [tex]\int_y^a f(t)dt= -\int_a^y f(t)dt[/tex]
     
  4. Mar 5, 2008 #3
    I understand that you couldn't follow it, but may I ask how to get the limits on the integral? Sorry for the mix-up.
     
  5. Dec 9, 2008 #4
    HallsofIvy...your d/dt should be a d/dx for the fund. theor.... the way you have it written it would be zero
     
  6. Dec 9, 2008 #5

    tiny-tim

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    Hey Kev! :smile:

    Hint: start by thinking ∫f = g …

    then ∂/∂x of (∫f between y and x)

    = ∂/dx ([g] between y and x). :wink:
     
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