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Math Help - Derivatives of Definite Integrals

  1. #1
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    Derivatives of Definite Integrals

    Given: f(x) = the integral from 0 to sin(x) of sqrt.(1 + t^2) dt
    and g(y) = the integral from 3 to y of f(x) dx.
    Find g''(pi/6)

    I'm not sure how to put it all together since there are two different integrals involved.
    Thanks!
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Derivatives of Definite Integrals

    in general

    \frac{\delta }{\delta  x}\int_{f(x)}^{g(x)} h(t) \cdot dt = h(g(x)) g'(x) - h(f(x)) f'(x)

    where did you stuck ?
    can you solve now
    can you write down the first derivative of g(y) ?
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  3. #3
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    Re: Derivatives of Definite Integrals

    Does g'(y) = f(y) ?
    If it does, then what do I do, since I only know what f(x) is?
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  4. #4
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    Re: Derivatives of Definite Integrals

    By the fundamental theorem of calculus:

    g'(y)=f(y)

    Then, g''(y)=f'(y). You just have to take the derivative of f(x). Use the Leibniz integral rule.
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  5. #5
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    Re: Derivatives of Definite Integrals

    Okay, so f'(x) = sqrt (1 + sin^2 (x) ), right?
    But I just don't quite understand how you get f'(y) when you only have f'(x). (I did try reading the Leibniz integral page, but I wasn't sure whether I was supposed to use it when differentiating f(x) or what.)
    Sorry to ask you to spell out every little step; this doesn't even seem like it should be such a hard problem, I'm just getting hung up with the switching between x and y.
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  6. #6
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    Re: Derivatives of Definite Integrals

    You shouldn't worry about the switching between x and y, that is just the name of the variable. The real question you have to answer is the derivative of f(x). Using the Leibniz integral rule should look someting like this:

    \begin{matrix}\frac{df}{dx}&=\frac{d}{dx}\int_0^{\  sin x}\sqrt{1+t^2}dt=\frac{d(\sin x)}{dx}.\sqrt{1+\sin^2 x}=\sqrt{1-\sin^4 x}\end{matrix}
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  7. #7
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    Re: Derivatives of Definite Integrals

    I think I've got it. Thanks a lot for all your help!
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