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Derivation with x^2 and root x

  1. Mar 19, 2007 #1

    disregardthat

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    1. The problem statement, all variables and given/known data

    derivate: [tex]x^2 \sqrt{x}[/tex]

    2. Relevant equations

    [tex](u \cdot v)^\prime = u^\prime \cdot v + v^\prime \cdot u[/tex]

    3. The attempt at a solution

    I found:
    [tex](x^2)^\prime = 2x[/tex]
    [tex](\sqrt{x})^\prime = \frac{1}{2 \sqrt{x}}[/tex]

    I entered them into the equation:

    [tex]2x \cdot \sqrt{x} + x^2 \cdot \frac{1}{2 \sqrt{x}} = 2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}}[/tex]

    But this is not correct!
     
  2. jcsd
  3. Mar 19, 2007 #2

    Hurkyl

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    Why do you think it is not correct?
     
  4. Mar 19, 2007 #3

    radou

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    Try to get rid of the square root in the denominator.
     
  5. Mar 19, 2007 #4

    robphy

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    Can you simplify your last expression?
     
  6. Mar 19, 2007 #5

    disregardthat

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    Ok, sorry. I checked it on my calculator many times and never got the correct result. But this time I did, by adding some extra brackets.. My calc just won't accept things with having it stuffed in with brackets!

    Simplify? I guess you could simplify it:
    [tex]2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}} = \frac{4x^{1.5}}{2} + \frac{x^{1.5}}{2} = \frac{5x^{1.5}}{2} = \frac{5x \sqrt{x}}{2}[/tex]

    Jesus christ, now my textbook's answer is correct too!
    Well, I guess that's a good thing. I must have seen on it uncorrectly. I got it right now... Much making a topic about ...
     
  7. Mar 19, 2007 #6

    Hurkyl

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    FYI, you can do this problem without the product rule -- do you see how?
     
  8. Mar 19, 2007 #7

    disregardthat

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    When you say it I think so, yes.

    [tex]x^2 \sqrt{x} = x^{2+0.5} = x^{2.5} = x^{\frac{5}{2}}[/tex]

    [tex](x^{\frac{5}{2}})^\prime = \frac{5}{2} \cdot x = \frac{5x}{2}[/tex]
     
  9. Mar 19, 2007 #8

    robphy

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    missing exponent?
     
  10. Mar 20, 2007 #9

    disregardthat

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    Of course
    [tex]\frac{5x^{\frac{3}{2}}}{2}[/tex]
     
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