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Math Help - derivative problem

  1. #1
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    derivative problem 2

    \[\frac{d}{{dx}}\left( {\,{x^{1/2}}\int {{x^{3/2}}f(x)dx} } \right)\]

    ????
    Last edited by mpx86; June 23rd 2013 at 08:48 AM.
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  2. #2
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    Re: derivative problem

    Hey mpx86.

    Hint: Use the product rule for differentiation and also use the chain rule for the fundamental theorem of calculus.
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  3. #3
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    Re: derivative problem

    I see how to apply the product rule and I can use the fundamental theorem of calculus on one of the terms. The other term becomes \frac{1}{2x^{1/2}}\int{x^{3/2}f(x)\,dx. Can it be simplified from there?

    - Hollywood
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  4. #4
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    Re: derivative problem

    I don't think that you can simplify in general any further.
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  5. #5
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    Re: derivative problem

    I have to ask about this one myself.

    After the product rule I have:

    \frac{1}{2\sqrt{x}}\int x^{\frac{3}{2}}f(x)dx + \sqrt{x}\frac{d}{dx}\int x^{\frac{3}{2}}f(x)dx

    I don't see how you would differentiate the integral in the second term with the FTOC because it is indefinite, the FTOC is always stated in terms of definite integrals as far as I understand it. I guess if say that \frac{d}{dx}\int g(x)dx=\frac{d}{dx}(G(x)+C)=g(x) is always true so I can say my x^{\frac{3}{2}}f(x) is my g(x) and just say \frac{d}{dx}\int x^{\frac{3}{2}}f(x)dx= x^{\frac{3}{2}}f(x)

    Or maybe I shouldn't ask if it is true, it is just definition of integral? Where does chain rule come in?
    Last edited by adkinsjr; June 24th 2013 at 04:52 PM.
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  6. #6
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    Re: derivative problem

    For some reason, I thought that the integral had limits that were a function of x.

    The post has been edited and I don't know if this was changed or not.
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  7. #7
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    Re: derivative problem

    Quote Originally Posted by chiro View Post
    For some reason, I thought that the integral had limits that were a function of x.

    The post has been edited and I don't know if this was changed or not.

    it wasnt edited for that reason but for a latex error........
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