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A simple computation using Leibniz's Rule

  1. Jan 22, 2012 #1
    A simple computational question, that I saw in a journal article and have been having trouble getting.

    Define the variable V (implicitly) by:

    [itex]V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s)[/itex],

    where r is a constant and F has support on [0,∞).

    Question: Show that [itex]\frac{dV}{db}=\frac{1}{1-rF(rV)}[/itex],

    My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get

    [itex]1=\frac{db}{dV} + r\cdot rVf(V) - 0 + \int^{rV}_0 r dF(s) + 0 - rVf(V) + 0 = \frac{db}{dV} + r^{2} Vf(V) + rF(rV) - rVf(V)[/itex]

    Rearrangement yields

    [itex]\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}[/itex]

    Notice that my solution has an additional ugly term in the denominator.

    Is my solution wrong? Or could perhaps a mistake have been made in the article?

    Thank you.
     
  2. jcsd
  3. Jan 25, 2012 #2
    bump =)
     
  4. Jan 25, 2012 #3

    chiro

    User Avatar
    Science Advisor

    Hey jakemf1986 and welcome to the forums.

    I think you have differentiated at least one of the integral terms incorrectly since one of the limits involves V. When this happens you need to use a special kind of chain rule.

    Does this help?
     
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