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

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: Leibniz Rule (derivative of an integral)
  1. Leibniz's rule (Replies: 5)