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Derivative and tangent line

  1. Oct 2, 2007 #1
    [SOLVED] Derivative and tangent line

    1. The problem statement, all variables and given/known data
    y = (x)/(x+x^-1)

    Find an equation of the tanget line to the graph at x=2


    2. Relevant equations



    3. The attempt at a solution

    I used the quotient rule and got the derivative to be 0 / (x+x^-1) = y' so then the slope is 0 of the tangent line; and I plugged x=2 into the original equation to find the corresponding y value which was 4/5 so my equation is just y=4/5 ? is this correct it seems odd and looks odd on my calculator
     
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  3. Oct 2, 2007 #2

    EnumaElish

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    How did you get 0 / (x+x^-1) = y' ?
     
  4. Oct 2, 2007 #3
    I got [(x+x^-1)-x(1-x^-2) ]/ (x+x^-1)^2 then disturbted the 2nd x and got (x+x^-1)-(x+x^-1) on the numerator but I think i just found my error; when distributing I turned the x^-2 in the 2nd term to a x^-1 but it should go to a x^-3 right?
     
  5. Oct 2, 2007 #4

    EnumaElish

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    No x^-1 is right.
     
  6. Oct 2, 2007 #5
    is the derivative (2x^-1)/(x+x^-1)^2
     
  7. Oct 2, 2007 #6

    EnumaElish

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    I looked at your solution and I think you solved it right the first time. What made you to doubt it?
     
  8. Oct 2, 2007 #7
    My friend who is working on the same problem got a different answer, and when I graphed the tangent line with that slope it looked like it crossed the graph at 2 points.
     
  9. Oct 2, 2007 #8

    EnumaElish

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    It is okay if it is tangent to the function at the specified point. It doesn't matter how many times it crosses the function at several points. Do you understand the difference between "being tangent" and "crossing"?
     
  10. Oct 2, 2007 #9
    I think i had a sign error my first time around and i made the 2nd x^-1 a negative when it should be positive then you have (x +x^-1 - x + x^-1)/(x+x^-1)^2 shouldn't the +x and -x negate each other but leave you with 2x^-1 for the numerator ?
     
  11. Oct 2, 2007 #10
    ohhh I thought that a tangent line could only hit the graph at one point
     
  12. Oct 2, 2007 #11

    EnumaElish

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    Why would it be +? Isn't (1/x)' = -1/x^2?
     
  13. Oct 2, 2007 #12
    yes it is but then the original derivative is [(x+x^-1) - x(1-x^-2)]/(x+x^-1)^2 so shouldn't the -x turn the x^-2 to +x^-1
     
  14. Oct 2, 2007 #13

    EnumaElish

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    No.

    1. A line L can be tangent to a function F at multiple points.
    2. A line L can cross a function F at multiple points.
    3. Any combination of the two is possible.

    Example: y = 1 is tangent to sin(x) at multiple points. y = 1/2 crosses sin(x) at multiple points.
     
    Last edited: Oct 2, 2007
  15. Oct 2, 2007 #14
    oh wow I didn't know that thank you, but shouldn't that -x^-2 term turn to +x^-1
     
  16. Oct 2, 2007 #15

    EnumaElish

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    You're right, I missed it. One way to check is to write y = 1/(1 + 1/x^2) then differentiate it. You should get the same answer.
     
    Last edited: Oct 2, 2007
  17. Oct 2, 2007 #16
    heh happens to the best of us thanks for all your help tonight you really cleared up some confused concepts in my thinking
     
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