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anonymous

  • one year ago

I need help with differentiating the following equation using implicit differentiation: x^2 - 2xy + y^3 = 6

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  1. anonymous
    • one year ago
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    do this d/dx (x^2) d/dx(2xy) d/dx(y^3) d/dx(6)

  2. anonymous
    • one year ago
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    Right, so that gives me 2x - 2y - 2y'x + 3y^2y' = 0 correct?

  3. anonymous
    • one year ago
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    right. Solve for y' and you're done

  4. anonymous
    • one year ago
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    This is what I get when I solve for y' but apparently its incorrect... \[y' = \frac{ 2y - 2x }{ 3y^2 - 2x }\]

  5. anonymous
    • one year ago
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    You differentiated it right so it must be that you made an algebraic mistake somewhere when solving for y'. check again.

  6. amistre64
    • one year ago
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    the algebra seems fine

  7. anonymous
    • one year ago
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    Really? So do you think that's the correct answer?

  8. amistre64
    • one year ago
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    maybe the system they are using thinks it can be reduced

  9. anonymous
    • one year ago
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    Right... But I also put the equation into an online calculator and it gets ... \[y' = \frac{ 2x-2y }{ 2x - 3y^2 }\]

  10. amistre64
    • one year ago
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    you can factor a 2 out of the top, but i dont see much else for it

  11. anonymous
    • one year ago
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    I keep looking through my calculations and I still don't see anything wrong

  12. amistre64
    • one year ago
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    factoring out a -1/-1 is not that important

  13. amistre64
    • one year ago
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    2(x-y) for the top

  14. anonymous
    • one year ago
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    Does it change the answer when you factor out a -1/-1 ?

  15. amistre64
    • one year ago
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    it only changes form since -1/-1 = 1

  16. amistre64
    • one year ago
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    (a-b) = -(b-a)

  17. anonymous
    • one year ago
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    Right, so it wouldn't be a necessary step to take right?

  18. amistre64
    • one year ago
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    its not required to me

  19. amistre64
    • one year ago
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    x^2 - 2xy + y^3 = 6 2x x' - (2x'y+2xy') + 3y^2 y' = 0 x' = dx/dx = 1 2x - 2y -2x y' + 3y^2 y' = 0 2x - 2y + (3y^2-2x) y' = 0 2(x -y) + (3y^2-2x) y' = 0

  20. amistre64
    • one year ago
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    y' = 2(y-x)/(3y^2 - 2x)

  21. anonymous
    • one year ago
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    That's exactly what I did! So I don't think it should be wrong

  22. amistre64
    • one year ago
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    its not wrong according to what you have provided us. can you give a screen shot or picture?

  23. anonymous
    • one year ago
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    Of the question?

  24. amistre64
    • one year ago
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    perferably yes ...

  25. anonymous
    • one year ago
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    Sure

  26. amistre64
    • one year ago
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    and how many tries do you have to get it 'correct' ?

  27. anonymous
    • one year ago
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    It's unlimited

  28. amistre64
    • one year ago
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    oh that good lol

  29. anonymous
    • one year ago
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    Haha i know right xD

  30. amistre64
    • one year ago
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    screen shot it when you input 2(y-x)/(3y^2 - 2x)

  31. anonymous
    • one year ago
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    kk hold on

  32. anonymous
    • one year ago
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    Oh wow now its saying that it's correct >.< This is ridiculous lol

  33. amistre64
    • one year ago
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    i was going to see if you were missing a paranthesis :)

  34. anonymous
    • one year ago
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    I'm always cautious of the parentheses XD there was something wrong with the program, I'm sure :D lol

  35. anonymous
    • one year ago
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    Thanks so much though!

  36. amistre64
    • one year ago
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    good luck :)

  37. anonymous
    • one year ago
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    Thanks! :)

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