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

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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..

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@hartnn

hartnn

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you'll need product rule for 3axy

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I thought product rule was used for function times function

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thats a variable,isn't it

hartnn

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y is the function of x, x is the function of x.

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:O

hartnn

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(xy)' = x'y+xy' = y+xy'

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so how will we find for 3axy?

hartnn

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3a (y+xy')

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\[\LARGE 3ay(\frac{dx}{dx}) |||| 3ax(\frac {dy}{dx}) \]

hartnn

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dx/dx=1

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yeah!

hartnn

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and i assume ||| is +

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yes

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thats it?:O

hartnn

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so, dy/dx will change..

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what do u mean?

hartnn

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i mean recalculate dy/dx

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\[\LARGE \frac{dy}{dx}= \frac{-3(ay+x^{2})}{3(y^{2}-ax)} \]

hartnn

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-ax ?

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sorry +
printing mistake :P

hartnn

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then ok.

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:D