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DLS

  • 2 years ago

x^3+y^3+3axy=0 dy/dx=?

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  1. DLS
    • 2 years ago
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    \[\LARGE 3x^{2}+3y^{2} \frac{dy}{dx}-3ax=0\] this gives..finally.. \[\LARGE \frac{dy}{dx}=\frac{x(a-x)}{y^{2}} \] I doubt this because im quite unsure about the derivative of 3axy,we are differentiating wrt x or y? if its x then 3ay(dx//dx) not sure..

  2. DLS
    • 2 years ago
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    @hartnn

  3. hartnn
    • 2 years ago
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    you'll need product rule for 3axy

  4. DLS
    • 2 years ago
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    I thought product rule was used for function times function

  5. DLS
    • 2 years ago
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    thats a variable,isn't it

  6. hartnn
    • 2 years ago
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    y is the function of x, x is the function of x.

  7. DLS
    • 2 years ago
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    :O

  8. hartnn
    • 2 years ago
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    (xy)' = x'y+xy' = y+xy'

  9. DLS
    • 2 years ago
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    so how will we find for 3axy?

  10. hartnn
    • 2 years ago
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    3a (y+xy')

  11. DLS
    • 2 years ago
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    \[\LARGE 3ay(\frac{dx}{dx}) |||| 3ax(\frac {dy}{dx}) \]

  12. hartnn
    • 2 years ago
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    dx/dx=1

  13. DLS
    • 2 years ago
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    yeah!

  14. hartnn
    • 2 years ago
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    and i assume ||| is +

  15. DLS
    • 2 years ago
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    yes

  16. DLS
    • 2 years ago
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    thats it?:O

  17. hartnn
    • 2 years ago
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    so, dy/dx will change..

  18. DLS
    • 2 years ago
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    what do u mean?

  19. hartnn
    • 2 years ago
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    i mean recalculate dy/dx

  20. DLS
    • 2 years ago
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    \[\LARGE \frac{dy}{dx}= \frac{-3(ay+x^{2})}{3(y^{2}-ax)} \]

  21. hartnn
    • 2 years ago
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    -ax ?

  22. DLS
    • 2 years ago
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    sorry + printing mistake :P

  23. hartnn
    • 2 years ago
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    then ok.

  24. DLS
    • 2 years ago
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    :D

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