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Derivative of a integral function?

  1. Jul 1, 2010 #1
    How does one work the following?

    [tex]\frac{d}{dx}\int^x_y f(x,u)du[/tex]

    I know that (given certain assumptions about the function f)

    [tex]\frac{d}{dx}\int^x_y f(w,u)du=f(w,x)[/tex]

    and

    [tex]\frac{d}{dx}\int^c_y f(x,u)du=\int^c_y \frac{df}{dx}(x,u)du[/tex]

    but how do we put them together?
     
  2. jcsd
  3. Jul 1, 2010 #2
    [tex]
    \frac{d}{dx}\int^x_y f(x,u)du = \int^x_y \frac{d}{dx}f(x,u)du+f(x,x).
    [/tex]
     
  4. Jul 2, 2010 #3
    Thanks, Pere.
     
  5. Jul 2, 2010 #4

    HallsofIvy

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    Staff Emeritus
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    In general, Lagrange's formula:
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= f(x,\beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt[/tex]

    In this particular problem y is independent of both x and u and can be treated as a constant: dy/dx= 0.
     
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