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amy0799

  • one year ago

Find d2^y/dx^2 in terms of x and y.

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  1. amy0799
    • one year ago
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    \[x ^{2}y ^{2}-8x=2\]

  2. madhu.mukherjee.946
    • one year ago
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    2xy^2+x^2(2y)dy/dx-8=0

  3. amy0799
    • one year ago
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    I already have the first derivative figured out, I don't know how to find the second though

  4. madhu.mukherjee.946
    • one year ago
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    just change the sides take evrything to right side except x^2(2y)dy/dx

  5. amy0799
    • one year ago
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    do I just plug in x^2y^2−8x into the derivative of d2^y/dx^2?

  6. madhu.mukherjee.946
    • one year ago
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    nope u do like this |dw:1442671687613:dw|

  7. madhu.mukherjee.946
    • one year ago
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    do you understand

  8. amy0799
    • one year ago
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    umm not really, is that for dy/dx?

  9. amy0799
    • one year ago
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    how? I cant really read what you drew

  10. madhu.mukherjee.946
    • one year ago
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    okay both the drawings

  11. amy0799
    • one year ago
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    can you type out ur equations please?

  12. madhu.mukherjee.946
    • one year ago
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    wait

  13. madhu.mukherjee.946
    • one year ago
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    dy/dx=8/x^2y-2xy^2/x^22y dy/dx=4/x^2y-y/x

  14. madhu.mukherjee.946
    • one year ago
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    now do double derivative

  15. amy0799
    • one year ago
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    I got the first derivative to be \[\frac{ xy ^{2}-4 }{ x^{2}y }\]

  16. madhu.mukherjee.946
    • one year ago
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    nope i told the first one

  17. madhu.mukherjee.946
    • one year ago
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    this is not normal derivative this is implicit differentiation

  18. amy0799
    • one year ago
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    I know but when I did the implicit differentiation for x^2y^2−8x=2 I got that equation and it's different than yours

  19. madhu.mukherjee.946
    • one year ago
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    yeah i'm sorry dude you are correct

  20. madhu.mukherjee.946
    • one year ago
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    i'm so sorry actually i've simplified it a bit

  21. amy0799
    • one year ago
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    how did you simplify it?

  22. amy0799
    • one year ago
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    Hello?

  23. madhu.mukherjee.946
    • one year ago
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    forget that what you did is absolutely correct

  24. madhu.mukherjee.946
    • one year ago
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    but u made a mistake with the signs it will be 4-xy^2/x^2y

  25. amy0799
    • one year ago
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    oh ok. so I just find the derivative of that?

  26. madhu.mukherjee.946
    • one year ago
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    yes

  27. amy0799
    • one year ago
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    ok. ill show u what I get in a minute

  28. madhu.mukherjee.946
    • one year ago
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    no problem

  29. amy0799
    • one year ago
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    I got \[\frac{ x ^{2}y(-2xyy'-y ^{2})-(4-xy ^{2})(x ^{2}y'+2xy) }{ y ^{2}x ^{4} }\] correct so far?

  30. amy0799
    • one year ago
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    r u there?

  31. madhu.mukherjee.946
    • one year ago
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    yes

  32. amy0799
    • one year ago
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    is what I had right so far?

  33. madhu.mukherjee.946
    • one year ago
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    yeah

  34. madhu.mukherjee.946
    • one year ago
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    anything more

  35. amy0799
    • one year ago
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    well I still haven't found the 2nd derivative.

  36. amy0799
    • one year ago
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    @IrishBoy123 can u help?

  37. madhu.mukherjee.946
    • one year ago
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    dude this is the second derivative

  38. amy0799
    • one year ago
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    ik but there should be more to it

  39. madhu.mukherjee.946
    • one year ago
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    see if you are able to simplify it more but this is f9

  40. amy0799
    • one year ago
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    f9?

  41. madhu.mukherjee.946
    • one year ago
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    yeah i mean fine

  42. madhu.mukherjee.946
    • one year ago
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    this is correct

  43. amy0799
    • one year ago
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    don't I need to plug in 4-xy^2/x^2y into dy/dx?

  44. madhu.mukherjee.946
    • one year ago
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    yes

  45. amy0799
    • one year ago
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    I got \[\frac{ 2x^2 y^5-8xy^3-16}{ y}\] but that's the wrong answer when I summited it

  46. IrishBoy123
    • one year ago
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    your first deriv looks wrong way round for starters

  47. amy0799
    • one year ago
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    what is it suppose to be?

  48. IrishBoy123
    • one year ago
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    should be \[\color{red} - \frac{ xy ^{2}-4 }{ x^{2}y }\]

  49. madhu.mukherjee.946
    • one year ago
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    nope the first derivative done by amy is f9

  50. madhu.mukherjee.946
    • one year ago
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    i've checked that in wolfram alpha

  51. IrishBoy123
    • one year ago
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    you can then do it implicitly starting \[y' .x^2 .y = 4 - xy^2 \] and applying the triple product, see here |dw:1442677649641:dw| but, whatever way you do it, it's horrible.

  52. madhu.mukherjee.946
    • one year ago
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    ????????

  53. amy0799
    • one year ago
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    @IrishBoy123 how would I find the answer?

  54. madhu.mukherjee.946
    • one year ago
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    your second derivative is wrong

  55. madhu.mukherjee.946
    • one year ago
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    y^2(y-3xy'(x))/x^2

  56. IrishBoy123
    • one year ago
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    first deriv \[x ^{2}y ^{2}-8x=2\] \[2x y ^{2} + x^22yy'-8 = 0\] \[y' = \frac{8-2xy^2}{2x^2y} = ....\]ie wrong for second deriv you **might** go: \[ (x^2)(y)(y')=4 -x y ^{2} \] \[ (x^2)'(y)(y') +(x^2)(y)'(y')+(x^2)(y)(y')' = -y^2-2xyy'\] at some point you plug in for y'

  57. amy0799
    • one year ago
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    isn't that gonna be a big number when I plug in y'?

  58. IrishBoy123
    • one year ago
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    it's going to be horrible whatever way you do it. i have a feeling this could be a more organised way to do it, but if you are most of the way through, carry on as before however, do check that first derivative again or you will never get there

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