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Differntiation on Integrals

  1. May 3, 2008 #1
    How would you differntiate the following type of formula:


    d/dt [ e^(Integral from 0 to t [v(x+s,t-s) ds])f(x+t)]


    Note: I dont want to change variables, I know I can make it nicer but I would like to know how to do it directly.



    So far:

    = e^(d/dt[integral from 0 to t(v(x+s,t-s)ds)]) * f(x+t) + e^(integral from 0 to t[v(x+s,t-s)ds]) * d/dt(f(x+t))

    So I really just need help doing:

    d/dt [ Integral from 0 to t (v(x+s,t-s)ds)]

    = v(x+s,t-s) right? I'm not sure about this.
     
  2. jcsd
  3. May 3, 2008 #2

    Hurkyl

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    Er,

    [tex]\frac{d}{dx} e^{u(x)} \neq e^{\frac{d}{dx} u(x)[/tex]

    (usually)
     
  4. May 3, 2008 #3

    Hurkyl

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    Nope. Apply Leibniz's rule. Or derive it yourself by using limits.
     
  5. May 4, 2008 #4
    True, though I still need to find out d/dx[u(x)] or in my case what I listed above. Since d/dt(e^u(x,t)) = d/dt[u(x,t)]e^u(x,t)


    Leibniz's rule:

    Since I'm unsure v or d/dt are continuous on over the region does that mean I cannot pass the differentiation?

    Or can I still use:

    d/dt[Integral from a to b(f(x,t)dx)] =

    Integral from a to b[d/dt(f(x,t)dx)] + f(b,t)db/dt - f(a,t)da/dt
     
    Last edited: May 4, 2008
  6. May 4, 2008 #5
    So, I think I did it but I'm unsure and I would like you guys to confirm it if possible.

    d/dt[ int 0 to t ( V(x+s,t-s)ds )
    =
    Int 0 to t [ d/dt ( v(x+s,t-s) ) ] + v(x+t, 0)dt/dt - v(x,t)d0/dt
    =
    Into 0 to t [ d/dt ( v(x+s,t-s) ) + v(x+t,0)

    Assuming continuity over the region
     
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