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Calculus quotient rule problem

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    using the quotient rule, find the derivative of y = (2x-3)/(√(x^2-5)). Do not leave the answer in complex form.


    2. Relevant equations
    the quotient rule g(x)f'(x) - f(x)g'(x)
    ---------------------
    (g(x))^2

    3. The attempt at a solution

    1/2+2x^2+3x/(x^2-5)(√(x^2-5))
     
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2

    jbunniii

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    How can you simplify [itex]\sqrt{x^2}[/itex]?

    Edit: Also, please confirm that the denominator is [itex]\sqrt{x^2} -5[/itex] as you have written, and not [itex]\sqrt{x^2 - 5}[/itex]. Your attempt at a solution shows that you are not always using parentheses when needed.
     
  4. Oct 22, 2012 #3
    square root of x^2= x. The square is over the x^2 and the -5 sorry for the error. I wanted to check my answer or be guided through the problem if I am incorrect. Thanks.
    Your correct in the specification of the square root. It should be the square root of x^2-5 with the square root sign over the entire expression.
     
    Last edited: Oct 22, 2012
  5. Oct 22, 2012 #4

    Mark44

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    Do you mean (2x-3)/√(x^2-5)? That's different from what you wrote.
    This is what you wrote:
    $$ \frac{1}{2} + 2x^2 + \frac{3x}{x^2 - 5} (\sqrt{x^2} - 5)$$

    If that's not what you intended, please add parentheses to correct it.
     
  6. Oct 22, 2012 #5

    Mark44

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    No, not true. ##\sqrt{x^2} = |x|##
     
  7. Oct 22, 2012 #6
    oh ok yeah I miswrote, thanks. Yes that is valid.
     
  8. Oct 22, 2012 #7
    Mod note: To reduce confusion, I fixed the erroneous correction.

    my fault yes I accidentally wrote the nine.
     
    Last edited by a moderator: Oct 22, 2012
  9. Oct 22, 2012 #8

    jbunniii

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    OK, so it's
    [tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]
    So let's identify the pieces you need for the quotient rule. What are [itex]f(x)[/itex], [itex]g(x)[/itex], [itex]f'(x)[/itex], and [itex]g'(x)[/itex]?
     
  10. Oct 22, 2012 #9
    (x^2-5)^1/2=g(x); f(x)=2x-3 I understand how to plug everything in I just am having problems with simplifying I suppose.
     
  11. Oct 22, 2012 #10

    jbunniii

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    What are [itex]f'(x)[/itex] and [itex]g'(x)[/itex]?
     
  12. Oct 22, 2012 #11
    (x^2-5)^(1/2) *2 - 1/2(x^2-5)^(-1/2) *2x
    -------------------------------------------------
    x^2-5
     
  13. Oct 22, 2012 #12

    Mark44

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    This part --> -1/2(x^2-5)^(-1/2) *2x <-- is g'(x). You are missing f(x).
     
  14. Oct 22, 2012 #13
    f'(x)=2 g'(x)=1/2(x^2-5)^-1/2 *2x
     
  15. Oct 22, 2012 #14
    new answer -23/2 /(x^2-5)
     
  16. Oct 22, 2012 #15

    jbunniii

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    By the way, it's pretty easy to typeset your expressions properly using the "tex" feature, and it will make your expressions a lot easier to read. To see an example of how to do it, right-click on the following equation and select "show math as tex commands":

    [tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]

    You can then copy/paste and modify as needed. Then add "tex" and "/tex" tags to make it work properly. For example, the equation above is specified as follows:

    Code (Text):
    [tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]
     
  17. Oct 22, 2012 #16
    cool, thanks for the tip, would my answer happen to be correct?
     
  18. Oct 22, 2012 #17

    jbunniii

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    If we make the fix that Mark44 pointed out, then (prior to simplifying), the quotient formula gives you
    [tex]\frac{2(x^2 - 5)^{1/2} - (1/2)(2x)(2x-3)(x^2 - 5)^{-1/2}}{x^2 - 5}[/tex]
    Now if you can tell us step by step how you are simplifying this, we can tell you what you are doing wrong.
     
  19. Oct 22, 2012 #18

    jbunniii

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    Nope, unfortunately still wrong.
     
  20. Oct 22, 2012 #19
    using the chain rule shouldn't you have (1/2)(2x-3)(x^2-5)^-1/2 *2x?
     
  21. Oct 22, 2012 #20

    jbunniii

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    I edited the formula above - I accidentally left out the "2x" factor in the second term of the numerator.
     
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