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Homework Help: Basic Derivative

  1. Feb 22, 2006 #1
    I need to find dy/dx of [tex]\frac{x^2 - 2x}{\sqrt{x}}[/tex]

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


    Does this look right so far?
     
    Last edited: Feb 22, 2006
  2. jcsd
  3. Feb 22, 2006 #2
    that's what I got
     
  4. Feb 22, 2006 #3
    It is right, but can be simplified quite alot.
     
  5. Feb 22, 2006 #4
    Yeah, I was having trouble getting the simplified answer in my book so I wanted to check whether I was on the right track.

    I can't get past here:

    [tex]\frac{2x\sqrt{x} - 2\sqrt{x} - x^2 - 2x}{2x\sqrt{x}}[/tex]
     
  6. Feb 22, 2006 #5
    I could have made a mistake (very likely :smile:) but i got

    [tex]\frac{\frac{3}{2}x-1}{\sqrt{x}}[/tex]
     
  7. Feb 22, 2006 #6
    Book says: [tex]\frac{3x - 2}{2\sqrt{x}}[/tex]
     
  8. Feb 22, 2006 #7
    Is my algebra just really bad or are those different? :p
     
  9. Feb 22, 2006 #8
    It's the same as mine.
     
  10. Feb 22, 2006 #9
    This doesn't look right at all. What did you do to get here?
     
  11. Feb 22, 2006 #10
    I multiplied out the left two terms in the numerator, and stuck the term with the negative exponant in the denominator. I assume now from what you said above that I need to multiply the other terms by 2sqrt(x) if I want to do that. correct?
     
  12. Feb 22, 2006 #11
    woo, I got it nevermind! Thanks for the help.
     
  13. Feb 22, 2006 #12
    You can't just take the term with the negative exponent and move it to the denominator.

    [tex]\frac{A-\frac{B}{C}}{D}\neq\frac{A-B}{CD}[/tex]

    You have to make a common denominator for the numerator (if that made sense), i.e.

    [tex]\frac{A-\frac{B}{C}}{D}=\frac{AC-B}{CD}[/tex]
     
  14. Feb 22, 2006 #13
    Ok, good thing.
     
  15. Feb 22, 2006 #14
    Yeah, I saw that in posts 5-6. That was my mistake. Thanks again.
     
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