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Math Help - Differentiating an integral-defined function

  1. #1
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    Differentiating an integral-defined function

    hi, i have a problem with differentiating please help

    (I think) It is well known that for an integral-defined function, differentiating them with the upper limit(i.e. 'x') takes the integral away. (*) i.e.
    \frac{d}{dx}\int_{0}^{x}\ f(t) \ dt = f(x)


    The problem is, I came across a weird question today:
    determine \frac{d}{dx}\int_{0}^{x}\frac{e^{t}}{e^t+e^{2x}} \ dt

    Well, integrating \frac{e^{t}}{e^t+e^{2x}} first and differentiating yields the (probably) correct answer
    \frac{1+2e^x-{e^{2x}}}{(e^x+1)(e^{2x}+1)}

    But when i apply the rule (*) 'differentiation of integral-defined function', it gives a completely different result
    \frac{e^x}{e^x+e^{2x}}

    Can anybody tell why?
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  2. #2
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    Re: Differentiating an integral-defined function

    Oh, and as far as i can think, the constant e^{2x} seems to be the problem. the rule always holds when the integrated function does not contain terms related to the differentiation.(and only consists variables of the integration)

    But i cannot figure out why the constant mess with the rule
    Last edited by Physicsdochi; July 31st 2013 at 11:11 PM.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Differentiating an integral-defined function

    If I were going to evaluate this derivative, I would use the substitution:

    u=e^t+e^{2x}\,\therefore\,du=e^t\,dt

    and now we have:

    \frac{d}{dx}\int_{1+e^{2x}}^{e^x+e^{2x}}\frac{1}{u  }\,du=\frac{e^x+2e^{2x}}{e^x+e^{2x}}-\frac{2e^{2x}}{1+e^{2x}}=

    \frac{1+2e^{x}}{1+e^{x}}-\frac{2e^{2x}}{1+e^{2x}}=\frac{1+2e^x-e^{2x}}{\left(1+e^{x} \right)\left(1+e^{2x} \right)}
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  4. #4
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    Re: Differentiating an integral-defined function

    It's because, for your rule, the contribution from the lower limit is independent of x and so gets differentiated out.
    For your actual problem, the contribution from the lower limit is a function of x.
    For your actual problem, the rule does not apply.
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  5. #5
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    Re: Differentiating an integral-defined function

    Ohh.. indeed! Thanks!
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