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jamroz

using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6

  • one year ago
  • one year ago

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  1. hba
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    The easiest way to do this is to use implicit differentiation on it as it stands. You differentiate using the normal rules but you treat y as an unknown function of x. This means f(y) differentates to f '(y)*(dy/dx) by the chain rule. xy + x + y^2 = 6 First implicit differentiation gives [1*y + x*1*(dy/dx)] + 1 + 2y*(dy/dx) = 0 .....[from xy].....................[from y^2] You can rearrange this to give dy/dx = (function of x and y) Second implicit differentiation gives dy/dx + [1*(dy/dx) + x*(d2y/dx2)] + [2*(dy/dx)*(dy/dx) + 2y*(d2y/dx2)] = 0 ...................[from x*(dy/dx)]......................[from 2y*(dy/dx)] Now you replace each dy/dx by its function of x and y found above. Then rearrange it to give d2y/dx2 = (function of x and y).

    • one year ago
  2. jamroz
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    i don't understand how you split it up, i'm confused.... that honestly made no sense

    • one year ago
  3. jamroz
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    yay now my head hurts even worse

    • one year ago
  4. hba
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    How Much More Questions Do You Have ?

    • one year ago
  5. hba
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    Or Is This The Last Question ?

    • one year ago
  6. jamroz
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    this is my first question...

    • one year ago
  7. hba
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    How Does Your Head Hurt ?

    • one year ago
  8. jamroz
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    i've got about 16 more for this section, then 5 on a second section on it and a worksheet

    • one year ago
  9. jamroz
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    i'm a junior in highschool taking calculus.. this isn't making sense

    • one year ago
  10. hba
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    Oh Wow Genius :D

    • one year ago
  11. jamroz
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    obviously not if this isn't making sense. sorry it's just frustrating!

    • one year ago
  12. jamroz
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    and it's not making a bit of sense

    • one year ago
  13. hba
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    Well, (x^2)y + x(y^2) = 6 d/dx[(x^2)y + x(y^2)] = d/dx (6)

    • one year ago
  14. hba
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    2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0

    • one year ago
  15. jamroz
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    i'm lost stop

    • one year ago
  16. jamroz
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    how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?

    • one year ago
  17. hba
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    You dont know how to differentiate ?

    • one year ago
  18. jamroz
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    I do, but it's confusing me

    • one year ago
  19. jamroz
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    the d/dx of x^2y?

    • one year ago
  20. jamroz
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    you use the multiplication rule right? udv + vdu?

    • one year ago
  21. hba
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    \[x ^{2}y+xy^2=6\]

    • one year ago
  22. jamroz
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    x^2(1) + y(2x) = 2xy + x^2

    • one year ago
  23. hba
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    Yes I am Using Product Rule

    • one year ago
  24. jamroz
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    then how did you get 2xy then dy/dx x^2?

    • one year ago
  25. jamroz
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    or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y

    • one year ago
  26. hba
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    \[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]

    • one year ago
  27. hba
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    Now Use Product Rule

    • one year ago
  28. hba
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    \[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]

    • one year ago
  29. hba
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    Now Take The Terms Including dy/dx On One Side

    • one year ago
  30. hba
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    \[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]

    • one year ago
  31. hba
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    Now Take Common dy/dx

    • one year ago
  32. hba
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    \[dy/dx(x^2+2xy)=-y^2-2xy\]

    • one year ago
  33. jamroz
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    then divide?

    • one year ago
  34. hba
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    So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]

    • one year ago
  35. hba
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    ^ Now Thats Your Answer Got It :)

    • one year ago
  36. jamroz
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    question, when using the product rule

    • one year ago
  37. jamroz
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    how do you know which one to put dy/dx with

    • one year ago
  38. jamroz
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    like when we had x^2y and got dy/dx(x^2) + 2xy

    • one year ago
  39. jamroz
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    why'd we put dy/dx with x^2 and not 2xy?

    • one year ago
  40. hba
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    Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula

    • one year ago
  41. hba
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    |dw:1350260697079:dw|

    • one year ago
  42. jamroz
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    okay but then which one do i put the dy/dx with?

    • one year ago
  43. jamroz
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    like the next problem is y2 = (x+1)/(x-1)

    • one year ago
  44. jamroz
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    so would i do 2y (dy/dx) = vdu - udv/v^2

    • one year ago
  45. hba
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    Depends Completely On You

    • one year ago
  46. jamroz
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    so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2

    • one year ago
  47. hba
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    @Godsgirl Please Help Her Out I Gotta Leave- Thanks :)

    • one year ago
  48. Godsgirl
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    Hey so wht do u need help with

    • one year ago
  49. Godsgirl
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    Hello?

    • one year ago
  50. jamroz
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    I'm here. I have a question on where to put the dy/dx when solving

    • one year ago
  51. jamroz
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    right now i'm working on y^2 = x-1/x+1

    • one year ago
  52. jamroz
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    so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?

    • one year ago
  53. Godsgirl
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    right

    • one year ago
  54. jamroz
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    and the left side will equal 2y when i take dy/dx of it

    • one year ago
  55. jamroz
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    now where do i keep dy/dx, when solving?

    • one year ago
  56. Godsgirl
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    here i already solved it

    • one year ago
  57. jamroz
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    where do i put it? because i know i need to get it alone

    • one year ago
  58. Godsgirl
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    I sent it to u on ur messages

    • one year ago
  59. jamroz
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    that's not the question i'm asking about..

    • one year ago
  60. jamroz
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    the new one is

    • one year ago
  61. Godsgirl
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    which one are u asking about then

    • one year ago
  62. jamroz
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    y^2 = x-1/x+1

    • one year ago
  63. jamroz
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    i'm not trying to come off as rude, i'm sorry!

    • one year ago
  64. jamroz
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    just frustrated

    • one year ago
  65. hba
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    I am Back

    • one year ago
  66. hba
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    \[y^2 = (x-1) /(x+1)\]

    • one year ago
  67. hba
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    differentiate both sides with respect to x

    • one year ago
  68. Godsgirl
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    oh okay i was lost there

    • one year ago
  69. Godsgirl
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    i see where shes at now

    • one year ago
  70. hba
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    \[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]

    • one year ago
  71. hba
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    \[2y dy/dx = 2/(x+1)^2\]

    • one year ago
  72. hba
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    \[dy/dx = y / (x+1)^2\]

    • one year ago
  73. hba
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    @jamroz Got It :)

    • one year ago
  74. jamroz
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    so you keep dy/dx with the y on the left

    • one year ago
  75. jamroz
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    but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?

    • one year ago
  76. hba
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    Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)

    • one year ago
  77. jamroz
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    thank you so much!

    • one year ago
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