How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $x^{3} + y^{3} = 3 x y^{2}$ $x^3+y^3=3xy^2$ ?

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Answer 1 · 2016-11-17T09:26:14

$\frac{d y}{d x} = \frac{(y + x) (y - x)}{y (y - 2 x)}$ $(dy)/(dx)=((y+x)(y-x))/(y(y-2x))$

$\frac{d}{d x} (x^{3} + y^{3}) = \frac{d}{d x} (3 x y^{2})$ $d/(dx)(x^3+y^3)=d/(dx)(3xy^2)$

the RHS will need the product rule.

$3 x^{2} + 3 y^{2} \frac{d y}{d x} = 3 y^{2} + 6 x y \frac{d y}{d x}$ $3x^2+3y^2(dy)/(dx)=3y^2+6xy(dy)/(dx)$

rearrange for $\frac{d y}{d x}$ $(dy)/(dx)$ , and simplify the algebra.

$3 y^{2} \frac{d y}{d x} - 6 x y \frac{d y}{d x} = 3 y^{2} - 3 x^{2}$ $3y^2(dy)/(dx)-6xy(dy)/(dx)=3y^2-3x^2$

$\frac{d y}{d x} (3 y^{2} - 6 x y) = 3 (y^{2} - x^{2})$ $(dy)/(dx)(3y^2-6xy)=3(y^2-x^2)$

$\frac{d y}{d x} = \frac{3 (y + x) (y - x)}{3 y (y - 2 x)}$ $(dy)/(dx)=(3(y+x)(y-x))/(3y(y-2x))$

$\frac{d y}{d x} = \frac{(y + x) (y - x)}{y (y - 2 x)}$ $(dy)/(dx)=((y+x)(y-x))/(y(y-2x))$

How do you find dydx(dy)/(dx) given x3+y3=3xy2x^3+y^3=3xy^2?