Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof involving FTC to two variables

  1. Mar 28, 2010 #1
    I'm given the following:
    If [tex] \frac{\partial^2 F}{\partial x \partial y} = f(x,y)[/tex]
    \int \int_R f(x,y) dA = F(b,d) - F(a, d) - F(b, c) + F(a, c)

    Where R = [a,b] x [c,d]

    My question: by integrating the inner integral, we get:

    \int_a^b \frac{\partial^2 F}{\partial x \partial y} dx = \left[ \frac{\partial F}{\partial y}\right]_a^b

    But this result is F_y. I was expecting F_x. Why? Because:

    \frac{\partial^2 F}{\partial x \partial y} = F_xy

    Which means: we differentiate with respect to x, then with respect to y.
    The result above seemingly implies that we differentiated with
    respect to y, then with respect to x. Hence, integrating F_yx,
    gives F_y, is this not correct? Or do they apply fubini's theorem
    and Clairaut's theorem above?
    Last edited: Mar 28, 2010
  2. jcsd
  3. Mar 29, 2010 #2


    User Avatar
    Science Advisor

    As long as the derivatives are continuous, the to ways of differentiating a mixed second derivative are equal:
    [tex]\frac{\partial }{\partial x}\frac{\partial f}{\partial y}= \frac{\partial}{\partial y}\frac{\partial f}{\partial x}[/tex].

    By integrating [tex]\frac{\partial^2 f}{\partial x\partial y}[/tex] with respect to x, you've "undone" the differentiation with respect to x leaving [tex]\frac{\partial f}{\partial y}[/tex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook