Question

Find $\frac{dy}{dx}$ of $x^3+x^{2}y+x{y}^2+y^3=81$

Answer 1

$x^3+x^{2}y+x{y}^2+y^3=81$
Differentiating both sides w.r.t $x$ we get,
$\frac{d}{dx}(x^3+x^{2}y+x{y}^2+y^3)=\frac{d}{dx}\left(81\right)$
$\Rightarrow 3x^2+[y.2x+x^2.\frac{dy}{dx}]+[y^2.1+x.2y.\frac{dy}{dx}]+3y^2\frac{dy}{dx}=0$
$\Rightarrow (x^2+2xy+3y^2)\frac{dy}{dx}+(3x^2+2xy+y^2)=0$
$\therefore \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$