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calyne Group Title

Use implicit differentiation to find an equation of the tangent line to the curve at the curve at the given point: x^2 + y^2 = (2x^2 + 2y^2 - x)^2

  • 2 years ago
  • 2 years ago

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  1. calyne Group Title
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    ... @ (0,1/2)

    • 2 years ago
  2. calyne Group Title
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    not that i tmatters anyway i'm just a little confused regarding the differentiation here

    • 2 years ago
  3. Outkast3r09 Group Title
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    alright give me a second to remember lol

    • 2 years ago
  4. Outkast3r09 Group Title
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    alright i'll upload a picture on what to do

    • 2 years ago
  5. Outkast3r09 Group Title
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    and an example give me a sec

    • 2 years ago
  6. calyne Group Title
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    cool thanks

    • 2 years ago
  7. calyne Group Title
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    like i got as far as [4x+4y*y']/[4x+4y*y'-1] = 2x^2 + 2y^2 - x

    • 2 years ago
  8. Outkast3r09 Group Title
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    actualy i can't... i'll just type it out what my book says GUIDELINES FOR IMPLICIT DIFFERENTIATION 1. Differentiat both sides of the equatio with respects to x 2. collect al terms involve dy/dx on the left side of the equation and move all others to right 3.factor dy/dx out of the left side of the equation 4.solve for dy/dx by dividing both sides

    • 2 years ago
  9. Outkast3r09 Group Title
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    by factor

    • 2 years ago
  10. calyne Group Title
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    right righ tok so and

    • 2 years ago
  11. calyne Group Title
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    so what do i do from what i have am i good so far

    • 2 years ago
  12. Outkast3r09 Group Title
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    let me get a piece of paper lol

    • 2 years ago
  13. cococupcoffee Group Title
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    ama differentiate for u

    • 2 years ago
  14. calyne Group Title
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    thanks k cool coco

    • 2 years ago
  15. calyne Group Title
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    love it

    • 2 years ago
  16. Outkast3r09 Group Title
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    so is coco checking if your differentiation is right?

    • 2 years ago
  17. cococupcoffee Group Title
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    one question. just verifying... for the secon part, everyting is raised to teh 2nd power?

    • 2 years ago
  18. Outkast3r09 Group Title
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    i'm guessing

    • 2 years ago
  19. cococupcoffee Group Title
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    oh, ok

    • 2 years ago
  20. Outkast3r09 Group Title
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    yes

    • 2 years ago
  21. cococupcoffee Group Title
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    ouch

    • 2 years ago
  22. cococupcoffee Group Title
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    umm. i suck @ algebra... and i think i am going to mess up if i do this. but, i am still going to go ahead and square all that. can someone then tell me if i was right?

    • 2 years ago
  23. Outkast3r09 Group Title
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    yeah i got completely something else than he did

    • 2 years ago
  24. Outkast3r09 Group Title
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    the second is a chain and the first is just regular

    • 2 years ago
  25. Outkast3r09 Group Title
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    i got 2x+2yy'=2(4x+4yy'-1)

    • 2 years ago
  26. Outkast3r09 Group Title
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    =2x+2yy'= 8x+8yy'-2

    • 2 years ago
  27. cococupcoffee Group Title
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    if you do implicit differentiation, then don't you do (for example, the first part): 2xdx+2ydy... then get all teh dx's and dy's on one side and divide so that dy/dx

    • 2 years ago
  28. Outkast3r09 Group Title
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    yes

    • 2 years ago
  29. Outkast3r09 Group Title
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    \[2yy'-8yy'=6x-2\]

    • 2 years ago
  30. cococupcoffee Group Title
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    ahh... i get it.. am so stupid! i was wondering where you got the chain rule from

    • 2 years ago
  31. Outkast3r09 Group Title
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    yep

    • 2 years ago
  32. Outkast3r09 Group Title
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    then you can pull out y'

    • 2 years ago
  33. Outkast3r09 Group Title
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    i think it's\[\frac{-6x-2}{6y}\]

    • 2 years ago
  34. Outkast3r09 Group Title
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    the - is outside parenthesis

    • 2 years ago
  35. Outkast3r09 Group Title
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    \[-(6x-2)/6y\]

    • 2 years ago
  36. Outkast3r09 Group Title
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    it's been a while though lol

    • 2 years ago
  37. Outkast3r09 Group Title
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    blah wait i messed up lol

    • 2 years ago
  38. Outkast3r09 Group Title
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    let me do this all over agani

    • 2 years ago
  39. Outkast3r09 Group Title
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    i forgot something

    • 2 years ago
  40. Outkast3r09 Group Title
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    it doesn't help that i'm also trying to help two people

    • 2 years ago
  41. cococupcoffee Group Title
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    ok, sorry guys but i am lost lol. but i wanna know how it's done. so outkast, once you did the chain rule, did you multiply or what?

    • 2 years ago
  42. Outkast3r09 Group Title
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    my chain rule was messed up lol it should be \[2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]

    • 2 years ago
  43. cococupcoffee Group Title
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    yes that's what i got

    • 2 years ago
  44. cococupcoffee Group Title
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    then what did you do

    • 2 years ago
  45. Outkast3r09 Group Title
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    i'd say to combine the right side

    • 2 years ago
  46. Outkast3r09 Group Title
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    it's going to be ugly hahah

    • 2 years ago
  47. cococupcoffee Group Title
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    dive 2xdx + 2ydy with 4xdx+4ydy...?

    • 2 years ago
  48. cococupcoffee Group Title
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    divide*

    • 2 years ago
  49. Outkast3r09 Group Title
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    i think you'd have to combine the right side which is a pain

    • 2 years ago
  50. cococupcoffee Group Title
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    Hey, do you know is there anyway i can like mark this question so that i can come back and look @ it 2morrow?

    • 2 years ago
  51. cococupcoffee Group Title
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    no seriously?! ouch... calyne, your proff/teacher must be coldhearted to give you such a question.

    • 2 years ago
  52. Outkast3r09 Group Title
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    um i think it should still be in the top left forner under notifications

    • 2 years ago
  53. Outkast3r09 Group Title
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    it's not that bad tho

    • 2 years ago
  54. cococupcoffee Group Title
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    Oh, ok. thanks!

    • 2 years ago
  55. Outkast3r09 Group Title
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    \[2x+2yy'=16x^3+16y^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'+2x\] \[2yy'=16x^3+16^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'\]

    • 2 years ago
  56. calyne Group Title
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    ...

    • 2 years ago
  57. Outkast3r09 Group Title
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    did you get what me and coco got about my last when you used the chain rule

    • 2 years ago
  58. calyne Group Title
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    what

    • 2 years ago
  59. calyne Group Title
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    you use the chain rule yeah du^2/du * du/dx first thing overall ik

    • 2 years ago
  60. calyne Group Title
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    what you multiply that whole thing by dy/dx too

    • 2 years ago
  61. Outkast3r09 Group Title
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    so whta did you get after you take dy/dx of both sides

    • 2 years ago
  62. calyne Group Title
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    what did you get damn

    • 2 years ago
  63. Outkast3r09 Group Title
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    using the chain rule i got 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)

    • 2 years ago
  64. calyne Group Title
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    sorry i got 2x+2yy' = 2(2x^2+2y^2-x) (4x + 4yy' - 1)

    • 2 years ago
  65. calyne Group Title
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    yeahhhh

    • 2 years ago
  66. Outkast3r09 Group Title
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    alright so if you combine the terms on the right

    • 2 years ago
  67. Outkast3r09 Group Title
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    is what i did ... what did you do after

    • 2 years ago
  68. calyne Group Title
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    idk that's gay

    • 2 years ago
  69. calyne Group Title
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    i tried a bunch of different stuff what was your final answer i'll tell you if it's correct nah flutter that the answer is x+1/2. is that what you got. if not we're wasting time.

    • 2 years ago
  70. Outkast3r09 Group Title
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    not done yet haha

    • 2 years ago
  71. Outkast3r09 Group Title
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    yeah i have no idea.. even wolfram alpha has omething like my end answer

    • 2 years ago
  72. Outkast3r09 Group Title
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    unless your teacher is crazy and solved for like y lol

    • 2 years ago
  73. Outkast3r09 Group Title
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    wait the answer to the differentiation is that or the tangent line

    • 2 years ago
  74. Outkast3r09 Group Title
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    because if it's the tangent line my implicit differentiation may be correct

    • 2 years ago
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