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amorfide

  • 3 years ago

okay so just out of curiousity if i had x² + xy + y² = 1 how would i differentiate?

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  1. hartnn
    • 3 years ago
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    heard of implicit differentiation ?

  2. amorfide
    • 3 years ago
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    no i have never heard of it

  3. Kainui
    • 3 years ago
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    you really just use the chain rule. Most of the time the derivative with respect to x is 1, so you don't think about the derivative of 3x^2 being 3x*1, but when you do implicit differentiation you take the derivative of 3y^2 to be 3y*(dy/dx)

  4. hartnn
    • 3 years ago
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    u want to diff w..r.t x, right ?

  5. Kainui
    • 3 years ago
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    6* not 3x or 3y by the way! lol

  6. amorfide
    • 3 years ago
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    yeah rtx

  7. hartnn
    • 3 years ago
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    ok, so u know product rule, and chain rule, can u differentiate xy ?

  8. amorfide
    • 3 years ago
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    no i can not lol i have never had xy together only y=x or something

  9. hartnn
    • 3 years ago
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    but u know product rule right ? take x and y as different functions and diff. xy using product rule

  10. hartnn
    • 3 years ago
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    like x'y+xy' and x' = ?

  11. amorfide
    • 3 years ago
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    would it be y+x?

  12. hartnn
    • 3 years ago
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    umm.....x'=1 right ? so diff. of xy will be y+xy' ok

  13. amorfide
    • 3 years ago
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    but isnt y' also 1? so it would be y+x?

  14. hartnn
    • 3 years ago
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    y' would have been 1 if you would have diff. w.r.t y as in d/dy of y =1 but here you are differentiating w.r.t x do its diff. would be dy/dx that is y' makes sense ?

  15. amorfide
    • 3 years ago
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    ohhh okay! thank you, that makes more sense

  16. hartnn
    • 3 years ago
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    and what will be diff. of y^2 then ?

  17. amorfide
    • 3 years ago
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    2y²' ?

  18. hartnn
    • 3 years ago
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    since d/dx of x^2 = 2x d/dy of y^2 = 2y dy/dx = 2yy' ok?

  19. amorfide
    • 3 years ago
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    okay so the diff of y^3 would be 3y²y' ?

  20. hartnn
    • 3 years ago
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    yup

  21. amorfide
    • 3 years ago
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    x² + xy + y² = 1 soo 2x+(x+y')+2y'=0 is that right

  22. hartnn
    • 3 years ago
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    2nd term ? y+xy'

  23. amorfide
    • 3 years ago
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    oops yeah that is what i meant 2x+(y+xy')+2y'=0

  24. amorfide
    • 3 years ago
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    2x+(y+xy')+2yy' =0

  25. hartnn
    • 3 years ago
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    yes, now find y' from there ?

  26. amorfide
    • 3 years ago
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    how do i do that? lol

  27. amorfide
    • 3 years ago
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    2x+(y+xy')+2yy' =0 -2x=y+xy' +2yy' -2x-y=xy'+2yy' -2x-y/2y=xy'/2y +y' uhm im stuck

  28. hartnn
    • 3 years ago
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    2x+(y+xy')+2yy' =0 xy'+2yy' = -2x-y (x+2y)y' = -(2x+y) now ?

  29. amorfide
    • 3 years ago
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    (x+2y)y' = -(2x+y) y'=-(2x+y)/(x+2y)

  30. hartnn
    • 3 years ago
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    yes :)

  31. hartnn
    • 3 years ago
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    wasn't it easy ?

  32. amorfide
    • 3 years ago
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    thank you! is that the entire thing finished?

  33. hartnn
    • 3 years ago
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    yup., u got your y'

  34. amorfide
    • 3 years ago
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    easier than i thought, thank you, i was just curious how to do a question like this, it isn't needed :P you are awesome!

  35. hartnn
    • 3 years ago
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    welcome ^_^

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