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

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  • DLS

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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  • DLS
\[\LARGE 3x^{2}+3y^{2} \frac{dy}{dx}-3ax=0\] 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..
  • DLS
you'll need product rule for 3axy

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Other answers:

  • DLS
I thought product rule was used for function times function
  • DLS
thats a variable,isn't it
y is the function of x, x is the function of x.
  • DLS
:O
(xy)' = x'y+xy' = y+xy'
  • DLS
so how will we find for 3axy?
3a (y+xy')
  • DLS
\[\LARGE 3ay(\frac{dx}{dx}) |||| 3ax(\frac {dy}{dx}) \]
dx/dx=1
  • DLS
yeah!
and i assume ||| is +
  • DLS
yes
  • DLS
thats it?:O
so, dy/dx will change..
  • DLS
what do u mean?
i mean recalculate dy/dx
  • DLS
\[\LARGE \frac{dy}{dx}= \frac{-3(ay+x^{2})}{3(y^{2}-ax)} \]
-ax ?
  • DLS
sorry + printing mistake :P
then ok.
  • DLS
:D

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