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

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


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

  4. May 3, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook