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mathcalculus Group Title

Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at the point (-3,1)?

  • 10 months ago
  • 10 months ago

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  1. hartnn Group Title
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    could you implicitly differentiate that ?

    • 10 months ago
  2. hartnn Group Title
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    you will need chain rule here whats d/dx x^2 =...? whats d/dx y^2=...?

    • 10 months ago
  3. hartnn Group Title
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    what ? where does 2*2 come from ?

    • 10 months ago
  4. hartnn Group Title
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) this is because of chain rule, got this step ?

    • 10 months ago
  5. hartnn Group Title
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    d/dx (x^2) = 2x thats why and the other 2 was already there

    • 10 months ago
  6. hartnn Group Title
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    ok, so now what about d/dx (x^2+y^2) =...?

    • 10 months ago
  7. mathcalculus Group Title
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    2x+2y?

    • 10 months ago
  8. hartnn Group Title
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    here, chain rule again comes into picture d/dx f(y) = f'(y) dy/dx so d/dx (y^2) = 2y dy/dx got this ?

    • 10 months ago
  9. hartnn Group Title
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    if you got that we are almost done :)

    • 10 months ago
  10. hartnn Group Title
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) this part ?

    • 10 months ago
  11. hartnn Group Title
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    whats d/dx x^2 =.... ?

    • 10 months ago
  12. hartnn Group Title
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    we don't rewrite anything

    • 10 months ago
  13. mathcalculus Group Title
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    ohman :/

    • 10 months ago
  14. hartnn Group Title
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    derivative of outer times derivative of inner like for (sin x)^3 we have outer = x^3, derivative = 3x^2 inner = sin x derivative = cos x so, 3 (sin x)^2 cos x got this ?

    • 10 months ago
  15. hartnn Group Title
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    in same way d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)

    • 10 months ago
  16. mathcalculus Group Title
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    i get that, but in this problem im confused with why the exponent goes away

    • 10 months ago
  17. hartnn Group Title
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    ok, you said d/dx (x^2) = 2x why did the exponent go away here ?

    • 10 months ago
  18. mathcalculus Group Title
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    2(x^2+y^2)^2=25(x^2-y^2)

    • 10 months ago
  19. mathcalculus Group Title
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    so you're finding the derivative of the inside?

    • 10 months ago
  20. hartnn Group Title
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    we do find derivative of inside in chain rule, right ?

    • 10 months ago
  21. mathcalculus Group Title
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    yup

    • 10 months ago
  22. mathcalculus Group Title
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    my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.

    • 10 months ago
  23. hartnn Group Title
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    he must have meant "derivative of outer first ....."

    • 10 months ago
  24. mathcalculus Group Title
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    d/dx[f(g(x))]= f' (g(x))*g'(x)

    • 10 months ago
  25. hartnn Group Title
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    yes, here f(x) was x2

    • 10 months ago
  26. mathcalculus Group Title
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    well isn't it 4?

    • 10 months ago
  27. mathcalculus Group Title
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    2(x^2+y^2)^2 |dw:1381774415672:dw|

    • 10 months ago
  28. hartnn Group Title
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    yes, it is 4.

    • 10 months ago
  29. mathcalculus Group Title
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    oh so then it is. where are we getting x2?

    • 10 months ago
  30. hartnn Group Title
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) = 4(x^2+y^2)times d/dx(x^2+y^2)

    • 10 months ago
  31. hartnn Group Title
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    sorry i didn't get your question?

    • 10 months ago
  32. mathcalculus Group Title
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    got that =]

    • 10 months ago
  33. mathcalculus Group Title
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    so can it be written like this? : d/dx 4(x^2+y^2) (2x+2y) ?

    • 10 months ago
  34. hartnn Group Title
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    no, why would d/dx come in front again ?

    • 10 months ago
  35. mathcalculus Group Title
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    so i get rid of it?

    • 10 months ago
  36. mathcalculus Group Title
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    since i have the derivative already?

    • 10 months ago
  37. hartnn Group Title
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    d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2) = 4 (x^2+y^2) times (2x+2y dy/dx) see whether you get this?

    • 10 months ago
  38. mathcalculus Group Title
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    yeah i got it, but why don't we do dy/dx here: 4 (x^2+y^2) times (2x+2y dy/dx)|dw:1381774790286:dw|

    • 10 months ago
  39. mathcalculus Group Title
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    we can't bc we have to leave it as it is right? according to the chain rule

    • 10 months ago
  40. hartnn Group Title
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    because there, you have not differentiated x^2+y^2 yrt

    • 10 months ago
  41. hartnn Group Title
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    *yet

    • 10 months ago
  42. hartnn Group Title
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    and yes, we need to leave it as it is, that too

    • 10 months ago
  43. mathcalculus Group Title
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    ok

    • 10 months ago
  44. mathcalculus Group Title
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    now what do we do

    • 10 months ago
  45. mathcalculus Group Title
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    same for the other side?

    • 10 months ago
  46. hartnn Group Title
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    yes, other side is too simple. try ?

    • 10 months ago
  47. hartnn Group Title
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    right side is just 25(2x-2y dy/dx) isn't it ?

    • 10 months ago
  48. mathcalculus Group Title
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    25(x^2-y^2)*(2x-2y*dy/dx)

    • 10 months ago
  49. hartnn Group Title
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    what why ? we need chain rule for just y^2 here, isn't it ?

    • 10 months ago
  50. mathcalculus Group Title
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    huh i did?

    • 10 months ago
  51. hartnn Group Title
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    its like 25x^2 - 25y^2 derivative = 25(2x) - 25 (2y dy/dx)

    • 10 months ago
  52. mathcalculus Group Title
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    where did i go wrong?

    • 10 months ago
  53. hartnn Group Title
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    the x^2-y^2 part was unnecessary, you though it was chain rule again ?

    • 10 months ago
  54. mathcalculus Group Title
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    wait it wasn't because it had no exponent?

    • 10 months ago
  55. hartnn Group Title
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    yes, correct.

    • 10 months ago
  56. mathcalculus Group Title
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    25*(2x-2y*dy/dx)

    • 10 months ago
  57. mathcalculus Group Title
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    so its this?

    • 10 months ago
  58. mathcalculus Group Title
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    got cha.

    • 10 months ago
  59. mathcalculus Group Title
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    now do we plug in x and y?

    • 10 months ago
  60. hartnn Group Title
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    correct! put x=-3, y=1 in that...

    • 10 months ago
  61. hartnn Group Title
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    i assume you are simplifying...

    • 10 months ago
  62. mathcalculus Group Title
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    i don't know i keep getting the answer wrong whenever i try to submit it

    • 10 months ago
  63. hartnn Group Title
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    what did u get for dy/dx ?

    • 10 months ago
  64. mathcalculus Group Title
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    i keep getting incorrects :/

    • 10 months ago
  65. mathcalculus Group Title
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    i got 3/119

    • 10 months ago
  66. hartnn Group Title
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    4 (x^2+y^2) times (2x+2y dy/dx) = 25(2x-2y dy/dx) x=-3, y=1 4(9+1) (-6 +2 dy/dx) = 25 (-6 -2 dy/dx)

    • 10 months ago
  67. mathcalculus Group Title
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    ohi wrote one of the signs wrong.

    • 10 months ago
  68. mathcalculus Group Title
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    got it.

    • 10 months ago
  69. mathcalculus Group Title
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    thank you so much!

    • 10 months ago
  70. mathcalculus Group Title
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    :)

    • 10 months ago
  71. hartnn Group Title
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    welcome ^_^

    • 10 months ago
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