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Differentiate a function using the quotient rule

  1. Jul 20, 2012 #1
    1. Differentiate



    2. y = [itex]\frac{v^{3} - 2v\sqrt{v}}{v}[/itex]



    3. I am trying to use the quotient rule, but am having trouble understanding how to use the square root. Here's what I have [itex]\frac{v(3v^{2}-2*1/2\sqrt{v}) - v^{3}+2v*1/2\sqrt{v}}{v^{2}}[/itex]
     
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  3. Jul 20, 2012 #2
    Remember that the square root of a number is equivalent to that number to the one-half power. So, [tex] \sqrt {x} = x^{\frac {1} {2}}[/tex] Replace the square roots with exponents, simplify the numerator and then use the power rule.
     
  4. Jul 20, 2012 #3

    eumyang

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    [itex]\sqrt{v} = v^{1/2}[/itex], so
    [itex]2v\sqrt{v} = 2v\;v^{1/2} = ?[/itex]


    EDIT: Beaten to it. :wink:
     
  5. Jul 20, 2012 #4
    I did, that's how I got [itex]\sqrt{v}[/itex] to equal [itex]\frac{1}{2\sqrt{v}}[/itex]. I am just having problems figuring out how to multiply it out.
     
  6. Jul 20, 2012 #5
    It equals v*v[itex]^{-1/2}[/itex]
     
  7. Jul 20, 2012 #6

    eumyang

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    Before you start differentiating, rewrite the expression without square roots:
    [tex]y = \frac{v^3 - 2v\;v^{1/2}}{v}[/tex]
    So what is
    [itex]2v\;v^{1/2}[/itex]?
     
  8. Jul 20, 2012 #7
    Is there a reason you are using the quotient rule?
     
  9. Jul 20, 2012 #8

    eumyang

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    If you were answering me, then I'm afraid you're wrong. Use the properties of exponents.
     
  10. Jul 20, 2012 #9

    eumyang

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    I was wondering the same thing, but I assumed that this is a problem from a textbook and the directions required the use of the quotient rule.
     
  11. Jul 20, 2012 #10
    That's what I was leaning toward - but I just want to be sure.
     
  12. Jul 20, 2012 #11
    It would be 2*[itex]\frac{3}{2}[/itex]v[itex]^{1/2}[/itex] which would equal 3v[itex]^{1/2}[/itex]
     
  13. Jul 20, 2012 #12

    eumyang

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    Well, that's the derivative of [itex]2v^{3/2}[/itex], but I was expecting you to say [itex]2v^{3/2}[/itex] first, before taking the derivative. So now the original expression becomes
    [tex]y = \frac{v^3 - 2v^{3/2}}{v}[/tex]
    Use the quotient rule from here.
     
  14. Jul 20, 2012 #13
    I see. Thanks for helping.
     
  15. Jul 20, 2012 #14
    I don't believe you calculated the derivative of the numerator correctly

    [itex] v^{3} - 2v \sqrt {v} [/itex]
    [itex] v^{3} - 2v v^{\frac {1} {2}} [/itex]
    [itex] v^{3} - 2v^{\frac {3} {2}} [/itex]
    Using the sum rule, differentiate both to get
    [itex] 3v^{2} - 3 \sqrt {v} [/itex]

    Use the quotient rule as you normally did, see if it's easier to simplify.

    EDIT: You beat me to it this time, you got your revenge.
     
  16. Jul 20, 2012 #15
    Can't you just simplify the equation and differentiate that?
     
  17. Jul 20, 2012 #16

    eumyang

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    On another forum (that also provided help in math), I recall that you could see who was viewing a particular thread. I remember that I would not bother replying to a thread if I saw that someone else was viewing the same thread.

    What do you mean? Are you now saying that we don't have to use the quotient rule? :confused:
     
  18. Jul 20, 2012 #17
    I'm just saying that wouldn't it be simpler to divide the numerator by v and differentiating the equation after that. But in the book it expects us to use the quotient rule.
     
  19. Jul 20, 2012 #18

    eumyang

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    Of course it would be simpler. I guess you haven't read all of the posts in this thread:

     
  20. Jul 20, 2012 #19
    Here's what I got using the quotient rule [itex]\frac{2v^{3}-v^{3/2}}{v^{2}}[/itex] which equals 2v -[itex]\frac{1}{\sqrt{v}}[/itex]
     
  21. Jul 20, 2012 #20

    eumyang

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    Correct.

    BTW, there's nothing wrong with putting an entire expression inside the tex tags. So instead of writing
    2v -[itex]\frac{1}{\sqrt{v}}[/itex]
    you could write
    [itex]2v - \frac{1}{\sqrt{v}}[/itex]
    Looks better, doesn't it? :smile:
     
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