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QUESTION

Find $\ds\frac{dy}{dx}$ at a general point on the curve
$y^3+xy+x^2=0$.

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ANSWER

$y^3+xy+x^2=0$ therefore $\ds
3y^2\frac{dy}{dx}+x\frac{dy}{dx}+y+2x=0$

i.e. $\ds\frac{dy}{dx}(3y^2+x)=-(y+2x).$ Therefore
$\ds\frac{dy}{dx}=-\frac{y+2x}{3y^2+x}$




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