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Thread: Differentiate a fraction

  1. #1
    Senior Member Mukilab's Avatar
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    Differentiate a fraction

    How would I differentiate this?

    $\displaystyle x^2+\frac{2000}{x}$

    I thought it would go like this (but it doesn't work). Where am I going wrong?::
    $\displaystyle x^2+2000*x^{-1}$>>>>>$\displaystyle x^2+2000^{-1}*x^{-2}=x^2+\frac{1}{2000}*\frac{1}{x^2}$

    Please help me out. Thank you in advance.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Differentiate a fraction

    $\displaystyle x^2+2000*x^{-1}>>>>>2* x - 2000*x^{-2}$
    Thanks from Mukilab
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Differentiate a fraction

    I would write the function as:

    $\displaystyle f(x)=x^2+2000x^{-1}$

    To differentiate, use the power rule term by term: $\displaystyle \frac{d}{dx}\left(kx^r \right)=krx^{r-1}$
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  4. #4
    Senior Member Mukilab's Avatar
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    Re: Differentiate a fraction

    Thanks guys
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  5. #5
    Senior Member Mukilab's Avatar
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    Re: Differentiate a fraction

    What something such as $\displaystyle \frac{24300x-225x^3}{16}$

    Surely since the 16^(-1) differentiates to 0, the whole equation differentiates to 0 (since everything is multiplied by 16^(-1))
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  6. #6
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    Re: Differentiate a fraction

    no ...

    $\displaystyle \frac{d}{dx} \left[ c \cdot f(x)\right] = c \cdot f'(x)$

    $\displaystyle y = \frac{1}{16}(24300x - 225x^3)$

    $\displaystyle y' = \frac{1}{16}(24300 - 675x^2)$


    fyi, problems involving derivatives belong in the calculus forum
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  7. #7
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Differentiate a fraction

    $\displaystyle f'(x)=\frac{1}{16} \left(24300-675 x^2\right)$
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  8. #8
    Senior Member Mukilab's Avatar
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    Re: Differentiate a fraction

    Quote Originally Posted by skeeter View Post
    no ...

    $\displaystyle \frac{d}{dx} \left[ c \cdot f(x)\right] = c \cdot f'(x)$

    $\displaystyle y = \frac{1}{16}(24300x - 225x^3)$

    $\displaystyle y' = \frac{1}{16}(24300 - 675x^2)$


    fyi, problems involving derivatives belong in the calculus forum
    Quote Originally Posted by MaxJasper View Post
    $\displaystyle f'(x)=\frac{1}{16} \left(24300-675 x^2\right)$
    Wow I was always told to multiply out before doing any differentiating. Thanks for that!
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  9. #9
    Senior Member Mukilab's Avatar
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    Re: Differentiate a fraction

    Can I do the same for integration? I.e. ignoring 15/16 until after I've integrated the thing in the brackets.
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  10. #10
    MHF Contributor
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    Re: Differentiate a fraction

    $\displaystyle \int k \cdot f(x) \, dx = k \int f(x) \, dx$


    look over your basic rules ... they're in your text.
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  11. #11
    Senior Member Mukilab's Avatar
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    Re: Differentiate a fraction

    Quote Originally Posted by skeeter View Post
    $\displaystyle \int k \cdot f(x) \, dx = k \int f(x) \, dx$


    look over your basic rules ... they're in your text.
    You would be surprised.

    How does one determine f(x)? Is it just anything involving an x? In other words, can I make anything k as long as it multiplies into the f(x) and does not involve x?
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  12. #12
    MHF Contributor
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    Re: Differentiate a fraction

    k is a constant factor of the function
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