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Math Help - Derivative of an Integral problem

  1. #1
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    Derivative of an Integral problem

    I need to find the derivative of an integral, with a limit of integration being a function:

    <br /> <br />
g(x) = \int_1^{x^{4}}(t^3+2)dt<br />

    How do I find g'(x)?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by AlderDragon View Post
    I need to find the derivative of an integral, with a limit of integration being a function:

    <br /> <br />
g(x) = \int_1^{x^{4}}(t^3+2)dt<br />

    How do I find g'(x)?
    Put:

    h(x)=\int_1^{x}(t^3+2)dt

    Then:

    g(x)=h(x^4)

    and so by the chain rule:

    \frac{d}{dx}g(x)=4x^3 \left. \frac{dh}{dx}\right|_{x^4}

    where the derivative of h is obtained by applying the fundamental theorem of calculus.

    CB
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  3. #3
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    Upon closer inspection of my past work I found another method of doing this type of problem. Substitute x^4 in for t and multiply by the derivative of x^4, 4x^3:

    <br /> <br />
g(x) = \int_1^{x^{4}}(t^3+2)dt<br />

    <br />
g'(x) = (x^{12} + 2)4x^3<br />

    Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by AlderDragon View Post
    Upon closer inspection of my past work I found another method of doing this type of problem. Substitute x^4 in for t and multiply by the derivative of x^4, 4x^3:

    <br /> <br />
g(x) = \int_1^{x^{4}}(t^3+2)dt<br />

    <br />
g'(x) = (x^{12} + 2)4x^3<br />

    Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?
    That is what you get if you finish what I posted in my last post, so yes it is correct. However why you do what you say you do (and what it means for that matter) needs explaining.

    CB
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  5. #5
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    By the way, here is the general Leibniz formula:
    \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= F(x, \beta(x))\frac{d\beta(x)}{dx}- F(x,\alpha(t))\frac{d\alpha(x)}{dx}+  \int_{\alpha(x)}^{\beta(x)}  \frac{\partial F(x,t)}{\partial x} dt.

    It can be proved from the fundamental theorem of calculus and using the chain rule for both the upper and lower limits.
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