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Homework Help: Defferentiate with respect to x

  1. May 23, 2010 #1
    2/x^(3) + 5x^(2/3)

    ok now i understand that this is what you do at the start,

    2x^(-3) + 5x^(2/3)
    Then Dy/Dx

    -6/x^(4) + (2/3)5x^(-1/3)

    After that i don't know what do do, the first part -6/x^(4) is fine whilst (2/3)5x^(-1/3) is meant to cancel down to

    10/3*root(X) but i got lost around here.
  2. jcsd
  3. May 23, 2010 #2


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    What exactly is your difficulty? I assume you know that (2/3)5= 10/3 so surely that is not the problem. Also you know that [itex]x^{-1/3}= 1/x^{1/3}[/itex] because you used that property to write [itex]2/x^3[/itex] as [itex]2x^{-3}[/itex].

    The only thing left is that [itex]x^{1/3}= \sqrt[3]{x}[/itex].

    [tex](2/3)5 x^{-1/3}= \frac{10}{3\sqrt[3]{x}}[/tex]

    The "root x" is a cube root, not a square root.
  4. May 23, 2010 #3
    HallsofIvy Thanks for the help i understand it a lot better now ^_^
    Last edited: May 23, 2010
  5. May 24, 2010 #4


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    Moderator's note: thread moved from Differential Equations. In the future please post homework questions, or any textbook-style questions that are for independent study, in one of the Homework & Coursework Questions forums.
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