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Partial Derivative

  1. Nov 7, 2003 #1
    I need the partial derivatives of:

    f(x,y) = ∫xy cos(t2) dt

    are they simply:

    ∂f/∂x = -2xcos(x2)
    and
    ∂f/∂y = 2ycos(y2)

    or am I completely lost here?
     
  2. jcsd
  3. Nov 7, 2003 #2

    NateTG

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    Are you sure it's not
    ∂f/∂x=-cos(x2)
    and
    ∂f/∂y=-cos(y2)

    ?
     
  4. Nov 7, 2003 #3
    No, I'm definitely not sure. Are you?
     
  5. Nov 7, 2003 #4

    NateTG

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    Well, lets say that
    ∫cos(t2) dt = g(t) + C
    then
    dg/dt= cos(t2) from the fundemental theorem of calculus.

    Now we have
    ∫xycos(t2) dt=g(y)-g(x)
    so I get
    ∂f/∂x=-cos(x2)
    ∂f/∂y=cos(y2)

    Does that make sense to you?
     
  6. Nov 8, 2003 #5
    I'm not sure. It's confusing.

    Isn't this a composite function that calls for use of the chain rule?

    I'm thinking that we have g(t) = t2 and the integral is f(g(t)) which gets evaluated at t=y and t=x, thus becoming f(g(x)) and f(g(y)), so cos(y2) - cos(x2) is df/dg and we still have to differentiate that wrt x to get ∂f/∂x and wrt y to get ∂f/∂y.

    So, I guess I'm saying
    ∂f/∂x = df/dg * ∂g/∂x [edited: I had the derivatives & the partials reversed]
    and
    ∂f/∂y = df/dg * ∂g/∂y [edited: same reason as above]

    Then,
    ∂f/∂x = ∂/∂x[cos(y2) - cos(x2)] = -2xcos(x2)
    and
    ∂f/∂y = ∂/∂y[cos(y2) - cos(x2)] = 2ycos(y2)

    But I'm not sure I have the f's and g's right, & maybe all of that is nonsense.
     
    Last edited: Nov 8, 2003
  7. Nov 8, 2003 #6

    HallsofIvy

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    Leibniz's formula is very general:

    d/dx(∫b(x)a(x)f(x,t)dt)=

    (db/dx)f(x,b(x))- (da/dx)f(x,a(x)+ ∫b(x)a(x)(∂f(x,t)/∂t dt)

    It doesn't matter that you are dealing with two variables, x, y, since with partial derivatives you are treating one of them as a constant.

    In particular, ∂/∂x(∫xycos(t2)dt= (-1)cos(x2) and
    ∂/∂y(∫xycos(t2)dt= (+1)cos(y2).
     
  8. Nov 8, 2003 #7

    NateTG

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    I think you're mixing your variables. You should probably use something other than f and x inside the integral.

    I also don't understand how Leibnitz's formula applies, since there is no multi-variable function inside the integral.

    As I stated above, all you need to do is apply the fundemental theorem of calculus.
     
  9. Nov 8, 2003 #8

    NateTG

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    gnome:

    Let's digress for a moment and look at the fundemental theorem of calculus:

    ∫xyg'(t)dt=g(y)-g(x)

    Now, let's say we have some g(t) so that g'(t)=cos(t2) so:

    ∫xycos(t2)dt=g(y)-g(x)

    Now, since g'(t)=cos(t2) we get
    ∂/∂x g(x) = g'(x)=cos(x2)
    and
    ∂/∂x g(y) = 0

    Does that make sense?
     
  10. Nov 8, 2003 #9
    Well, I'm sure you're right & I'm sitting here with my textbook from calc I opened to the Fundamental Theorem of Calculus trying to make sense of it. To simplify matters, I'm going to forget about partial derivatives for the moment, & just look at these:

    f(x) = cos(2x-1)
    f'(x) = -2sin(2x-1) (of this, I'm certain)

    -------------------------------------

    Now, what you're telling me is

    g(x) = ∫sin(2x-1)dx
    g'(x) = sin(2x-1)

    Right?

    And the more I think about that, the more sense it seems to make. But if you can add anything to that to make it clearer, please do.

    Thanks.
     
  11. Nov 8, 2003 #10

    NateTG

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    You've pretty much got it.
     
  12. Nov 12, 2003 #11

    HallsofIvy

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    No, I was not mixing variables. It was necessary to use f and x "inside the integral" because they appeared outside the intergral. The only dummy variable was t.

    Leibnitz's formula still applies, the last integral happens to be 0. The original question was about how you handled functions of x in the limits of integration. Leibnitz's formula does that nicely.
     
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