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Math Help - rational exponents

  1. #1
    DBA
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    rational exponents

    Hello, I try to understand a step in my solution manual, but I can not figure out what they did. Please help me with some steps they obvoiusly left out...

    x^4/5 * 2(x-4) + (x-4)^2 * 4/5x^-1/5 =

    1/5x^-1/5 *(x-4) *[5*x*2+(x-4)*4]


    How do they get from line 1 to line 2? Thanks for any help
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  2. #2
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    Quote Originally Posted by DBA View Post
    Hello, I try to understand a step in my solution manual, but I can not figure out what they did. Please help me with some steps they obvoiusly left out...

    x^4/5 * 2(x-4) + (x-4)^2 * 4/5x^-1/5 =

    1/5x^-1/5 *(x-4) *[5*x*2+(x-4)*4]


    How do they get from line 1 to line 2? Thanks for any help
    x^{\frac{4}{5}}\cdot 2(x-4) + (x-4)^2 \cdot \frac{4}{5}x^{-\frac{1}{5}} = \frac{1}{5}\cdot 5 \cdot x \cdot x^{-\frac{1}{5}} \cdot 2(x-4) +  (x-4)(x-4) \cdot 4 \cdot \frac{1}{5} \cdot x^{-\frac{1}{5}}

     = \frac{1}{5}\cdot x^{-\frac{1}{5}} \cdot (x-4) \cdot 5 \cdot x \cdot 2 + \frac{1}{5}\cdot x^{-\frac{1}{5}}\cdot (x-4)(x-4)\cdot 4 =  \frac{1}{5}x^{-\frac{1}{5}}(x-4)(5 \cdot x \cdot 2) + \frac{1}{5}x^{-\frac{1}{5}}(x-4)( 4(x-4) ) = \frac{1}{5}x^{-\frac{1}{5}}(x-4)(5\cdot x \cdot 2 + (x-4)4)
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  3. #3
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    Hello, DBA!

    I cannot figure out what they did.

    x^{\frac{4}{5}}\cdot2(x-4) \:+\: (x-4)^2 \cdot\frac{4}{5}x^{-\frac{1}{5}}

    . .   =\;\frac{1}{5}x^{-\frac{1}{5}}(x-4)\cdot\bigg[5\cdot x\cdot2+(x-4)]\cdot4\bigg]

    How do they get from line 1 to line 2?
    Is that the way they wrote it ? . . . How ugly!


    Evidently, this is a derivative of a product.

    y \:=\:x^{\frac{4}{5}}(x-4)^2

    Then: . \frac{dy}{dx} \;=\;x^{\frac{4}{5}}\cdot2(x-4) + (x-4)^2\cdot\frac{4}{5}x^{-\frac{1}{5}}

    . . . . . . . = \;2x^{\frac{4}{5}}(x-4) + \frac{4}{5}x^{-\frac{1}{5}}(x-4)^2 \;= \;2x^{\frac{4}{5}}(x-4) + \frac{4(x-4)^2}{5x^{\frac{1}{5}}}


    Get a common denominator: . \frac{2x^{\frac{4}{5}}(x-4)}{1}\cdot{\color{blue}\frac{5x^{\frac{1}{5}}}{5x  ^{\frac{1}{5}}}} + \frac{4(x-4)^2}{5x^{\frac{1}{5}}} . = \;\frac{10x(x-4)}{5x^{\frac{1}{5}}} + \frac{4(x-4)^2}{5x^{\frac{1}{5}}}


    . . = \;\frac{10x(x-4) + 4(x-4)^2}{5x^{\frac{1}{5}}} \;=\;\frac{2(x-4)\bigg[5x + 2(x-4)\bigg]}{5x^{\frac{1}{5}}} \;=\;\boxed{\frac{2(x-4)(7x-8)}{5x^{\frac{1}{5}}}}

    which can be written: . \frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)

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  4. #4
    DBA
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    Thanks for your help, I understand it now :-)
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  5. #5
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    Sorry. Mistake
    Last edited by Alienis Back; November 30th 2009 at 07:42 AM.
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