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Partial derivative of an Integral

  1. May 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Show
    [tex]
    \partial /\partial u \int_{a}^{u} f(x,v) dx = f(u,v)
    [/tex]

    2. Relevant equations



    3. The attempt at a solution
    Basically i understand that we hold all other variables constant, and i understand that we will get our answers as a function of u and v. But to show that we have f(u,v) im not too sure

    [tex]
    \int_{a}^{u} f(x,v)dx
    = g(u,v)- g(a,v)\\
    [/tex]

    [tex]
    \partial /\partial u [g(u,v)- g(a,v)]= g'_{u}(u,v)
    [/tex]
     
  2. jcsd
  3. May 31, 2009 #2
    You are correct in your formulas, but the dependency on two variables may be obfuscating your intuition. Note that unless v is a function of x, the expression:
    [tex]\frac{\partial}{\partial u} \int_{a}^{u} f(x,v) dx[/tex]
    is equivalent to the single-variable calculus expression:
    [tex]\frac{d}{du} \int_a^u f(x) dx[/tex]
    where we have suppressed the dependence of f on v because in this expression v is treated as a constant anyway. From here it is just the fundamental theorem of calculus, as you have applied. To be a little more rigorous, you can let gv(x) = f(x, v) and work on that function instead.
     
  4. May 31, 2009 #3
    heyy u know about newton leibnitz formula..?
    try to correlate ur prob wid this:
    [tex]\frac{d}{dx}[/tex][tex]\int_g^hf(x,t)dt[/tex] where g=g(x) and h=h(x)
    can be evaluated as:
    [tex]\frac{d}{dx}[/tex][tex]\int_g^hf(x,t)dt=[/tex] [tex]\int_g^h \frac{\partial}{\partial x}f(x,t)dt[/tex] [tex]+f(x,h(x)) \frac{d}{dx} h(x) -f(x,g(x)) \frac{d}{dx} g(x)[/tex]
    there is a very simple proof to it...which you may do yourself..(as i am a beginer at latex..so im in no mood to type more of programs..after this programming nightmare today)
    that should do it..the thing above is the most general case...
    you only have to replace the complete derivative with the partial one..wherever the need arises..
    xDD
     
    Last edited: May 31, 2009
  5. May 31, 2009 #4
    Thanks both,

    I think then if i let gu(x,v) = f(x, v)

    Then i will integrate as before to get

    [tex]

    \partial /\partial u [g(u,v)- g(a,v)]= g_{u}(u,v)

    [/tex]

    so following on we can say if
    gu(x,v) = f(x, v)
    then
    gu(u,v) = f(u, v)
    (since x is the dummy variable)

    Would this be correct?

    vaibhav1803, The context of the question which i did not state was to help in understanding a particular step in deriving leibniz, so i wanted to avoid actually using leibniz.
     
  6. May 31, 2009 #5
    i feel liebnitz is easier to do than assuming a new function..anyways
    the case i considered as i mentioned is the worst case possible..everytijng being a function..
    but for the partial outside differentiate taking x as the only variant
    yup x is a dummy variable...only the function affects the integral..thats correct
    glad if i cud help
     
  7. May 31, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You don't really need all of Leibniz's formula. Since the only depenence on u is the upper limit of integration, it's just the "fundamental theorem of calculus".
     
  8. May 31, 2009 #7
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