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Math Help - Finding the derivative of an integral

  1. #1
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    Finding the derivative of an integral

    The question:
    Find F'(x) for the function F:R -> R given below:
    F(x) = \int^{x^3}_0 {sin(t^2)dt

    My attempt:
    It is my understanding that the First Fundamental Theorem of Calculus states that:
    F'(x) = f(x)

    So my solution was just  sin(x^2). However, this isn't the correct solution. I'm not sure what I'm doing wrong. Any assistance would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:
    Find F'(x) for the function F:R -> R given below:
    F(x) = \int^{x^3}_0 {sin(t^2)dt

    My attempt:
    It is my understanding that the First Fundamental Theorem of Calculus states that:
    F'(x) = f(x)

    So my solution was just  sin(x^2). However, this isn't the correct solution. I'm not sure what I'm doing wrong. Any assistance would be appreciated. Thanks.
    if \displaystyle F(x) = \int_a^u f(t) \, dt where u is a function of x , then \displaystyle F'(x) = f(u) \cdot \frac{du}{dx}
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  3. #3
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    Thanks!
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  4. #4
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    More generally, Leibniz' rule:
    \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= f(x, \alpha(x))\frac{d\alpha}{dx}- f(x, \beta(x))\frac{d\beta}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}dt

    Which can be derived by the fundamental theorem of calculus and the chain rule as skeeter suggests.
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  5. #5
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    Interesting, Leibniz isn't mentioned in this text. Thanks!
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