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Leibniz Rule (derivative of an integral)

  1. Nov 3, 2004 #1
    The Leibniz rule:

    1. Let f(x,y) be a continuous two variable real function defined on (closed intervals) {x0, x1} x {y0, y1}.
    2. Let f_1 (partial derivative of f with respect to the first variable) exists and be continuous on the same subset of RxR.
    3. Let F be defined as F(x) = (int) (lim x = y0, x = y1) f(x,y)dy

    Then F' exists and F'(x) = (int) (lim x = y0, x = y1) f_1(x,y)dy.

    I know and understand the proof that uses the Mean Value Theorem of elementary calculus. Yet in Wikipedia (Leibniz's rule: derivatives and integrals) they use a simpler proof. They arrived to this point, which I understood:

    F'(y) = lim (h->0) (int)(lim x=c, x=d) ( f(x+h, y) - f(x,y)) dy

    BUt then they say: "... and using uniform continuity the right hand side equals to

    (int) (lim x=c, x=d) f_1(x,y)dy."

    How do they manage to use uniform continuity to introduce the limit inside the integral ????

  2. jcsd
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