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How can I differentiate this expression?

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider three univariate distinct functions [itex]f_1(x),f_2(y),f_3(y)[/itex]. Let H be given by the following integral:

    [itex] H=\int_{0}^{f_1(x)} G(f2(\xi))G(f3(\xi))d\xi [/itex]


    3. The attempt at a solution
    Then, computing dH/dy should give zero. However, I am not certain of this because the chain rule would give me:

    [itex] \frac{dH}{dy}=\frac{\partial H}{\partial y} + \frac{\partial H}{\partial f_2}\frac{\partial f_2}{\partial y}+ \frac{\partial H}{\partial f_3}\frac{\partial f_3}{\partial y}[/itex]

    and [itex]\frac{\partial H}{\partial f_j}\frac{\partial f_j}{\partial y}[/itex] are nonzero.
     
  2. jcsd
  3. Jan 21, 2012 #2

    SammyS

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    I had hoped one of our "experts" such as Dick, or micromass, or ... would have responded to this, but after looking at it a few times, I'll take a stab at it. (An informed stab at that.)

    The integrand [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] is a function of ξ. You are integrating over ξ and the limits of integration are 0 and f1(x). Therefore, H is a function of x and only of x, so of course, [itex]\displaystyle \frac{dH}{dy}=0\,.[/itex]

    If you intended the limits of integration to be 0 and f1(y), then finding [itex]\displaystyle \frac{dH}{dy}[/itex] makes sense.

    If [itex]\displaystyle H=\int_{0}^{f_1(y)} G(f_2(\xi))G(f_3(\xi))d\xi\,, [/itex] the H(y) is the anti-derivative of [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] evaluated at ξ=f1(y), minus a constant.

    The derivative of this result (w.r.t. y) should be pretty obvious, if this is indeed the question.
     
  4. Jan 22, 2012 #3
    Thanks SammyS. I thought exactly the same as you. However, I still don't see why the chain rule would be invalidated in this case. I know that so long as y does not appear in the limits of integration the integral should not change with y, but why the chain rule doesn't appear to say the sameÉ
     
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