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Derivative of (x^3+y^3)xy=(x^5-y^3)
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\[\large (x^3+y^3)xy=(x^5-y^3) \]
Solve for dy/dx
@zepdrix
\[\large (x^4y+xy^4)=(x^5-y^3) \] like u said^^
\[\large 4x^3y+x^4 \frac{dy}{dx}= x^4-3y^2 \frac{dy}{dx}\]
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\[\frac{dy}{dx}(x^4+3y^2)=x^4-4x^3y\]
\[\large \frac{dy}{dx}=\frac{x^4-4x^3y}{x^4+3y^2} =\frac{x^3(x-4y)}{x^4+3y^2}\]
sup sensei :D
\[\large \left(4x^3y+x^4 \frac{dy}{dx}\right) + \qquad \qquad \qquad = 5x^4-3y^2 \frac{dy}{dx}\] Hmm it looks like you applied the product rule to the first term, what about the other one? :O
ive applied it,check the next steps!
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@zepdrix ?? :(
I dunno, I can't follow your steps :\
okay..lets move!
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