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Deriving a root and a fraction

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Derive [itex]y= \sqrt(2x^4) - \frac{5}{3x^2}[/itex]



    3. The attempt at a solution

    I am still at the first part of the function (the root):
    First I tried to derive inside the root like this:
    [tex] \sqrt(2x^4) = \sqrt(8x^3) = 2 \sqrt(2) \times \sqrt(x^3) = \sqrt(2) \times x^2[/tex]

    Unfortunately, the first part of the function is supposed to be 2\sqrt(2)x.

    The second part I am nowhere close yet.
     
  2. jcsd
  3. Oct 16, 2011 #2

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    Hi Twinflower! :smile:

    Before you take the derivative, you should first try to simplify your expression.

    Did you know that [itex]\sqrt {(2x)} = \sqrt 2 \sqrt x[/itex]?
    And that [itex]\sqrt {(x^4)} = x^2[/itex]?
     
    Last edited: Oct 16, 2011
  4. Oct 16, 2011 #3
    Hi, thanks for your reply.

    I tried what you are suggesting, but I somehow ended up with [itex]\sqrt(x^4) = x^3[/itex] since [itex]\sqrt(x^2) = x^1 = x[/itex]

    Second attempt:

    [tex]\sqrt(2x^4)[/tex]
    [tex]\sqrt(2) \times \sqrt(x^4)[/tex]
    [tex]\sqrt(2) \times x^2[/tex]
    [tex]2\times \sqrt(2) \times x[/tex]
    [tex]2 x[/tex]

    hm.... didn't go quite well now either?
     
  5. Oct 16, 2011 #4

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    Actually, your second attempt is looking quite well. :)


    But how did you get from:

    [itex]2\times \sqrt(2) \times x[/itex]

    to

    [itex]2 x[/itex]?
     
  6. Oct 16, 2011 #5
    Oh, my bad.
    Brain-fart.

    For a second I though that 2 times sqrt2 equals 2.

    The final answer should ofcourse be
    [tex]2 \times \sqrt(2) \times x[/tex]
     
  7. Oct 16, 2011 #6
    Thank you by the way.

    Now I need a push in the right direction regarding the fraction :)
     
  8. Oct 16, 2011 #7

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    You're welcome! :smile:


    For the fraction you need to know that for instance [itex]{1 \over x^5} = x^{-5}[/itex].

    Oh, and also that [itex]{2 \over 3x} = {2\over 3} \times {1 \over x}[/itex].
     
  9. Oct 16, 2011 #8
    Ah, of course!
    Gimme a minute, and I'll figure that one out as well :)
     
  10. Oct 16, 2011 #9
    Ok, here we go:


    [tex]- \frac{5}{3x^2}[/tex]
    [tex]- \frac{5}{3} \times \frac{1}{x^2}[/tex]
    [tex]- \frac{5}{3} \times x^{-2}[/tex]
    [tex]- \frac{5}{3} \times -2 \times x^{-3}[/tex]
    [tex]\frac{10}{3} \times x^{-3}[/tex]
    [tex]\frac{10\times x^{-3}}{3}[/tex]
     
  11. Oct 16, 2011 #10
    Thanks a bunch! :)
     
  12. Oct 16, 2011 #11

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    Hey! Your minute is up!! :wink:

    Edit: errrr.... I guess you were just in time! :rolleyes:
     
  13. Oct 16, 2011 #12
    yeye, I needed to write it down in my exercise paper and THEN i had to write it all in LaTeX (which is still quite messy to me)

    :)
     
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