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Math Help - power derivative

  1. #1
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    power derivative

    Hi,

    I was wondering how I would convert  \frac{1}{3x^3} and  \frac{1}{3x} to "power form". For example \frac{2}{x^3} would be 2x^-3 but when I try the same logic with the first two,  \frac{1}{3x^3} would become  {3x^-3} but that's wrong? When 1 is the numberator I seem to get them wrong.
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  2. #2
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    Quote Originally Posted by Ifailatmaths View Post
    Hi,

    I was wondering how I would convert  \frac{1}{3x^3} and  \frac{1}{3x} to "power form". For example \frac{2}{x^3} would be 2x^-3 but when I try the same logic with the first two,  \frac{1}{3x^3} would become  {3x^-3} but that's wrong? When 1 is the numberator I seem to get them wrong.
    The constants stay where they are in the fraction.

    In your example:

     \frac{1}{3x^2} = \frac{1}{3} \times \frac{1}{x^2}

    The law of indices states that  \frac{1}{a^b} = a^{-b}

    Hence we get:


     \frac{1}{3} \times \frac{1}{x^2} = \frac{1}{3} \times x^{-2} = \frac{x^{-2}}{3}
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  3. #3
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    Thanks mush but Im trying to get it into the form so I can differentiate and for that don't I need it without any fractions and just powers?
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  4. #4
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    Quote Originally Posted by Ifailatmaths View Post
    Thanks mush but Im trying to get it into the form so I can differentiate and for that don't I need it without any fractions and just powers?
    Nope. When you differentiate a term  \frac{p}{q} x^{n} where p and q are constants, then the constants come OUTSIDE the integration.

     \frac{d}{dx} (\frac{x^{-2}}{3})

     =\frac{d}{dx} (\frac{1}{3} \times x^{-2})

     =\frac{1}{3} \times \frac{d}{dx} (x^{-2})

     =\frac{1}{3} \times (-2x^{-3})

     =\frac{-2x^{-3}}{3}

     =\frac{-2}{3x^{3}}
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