Question

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

Answer 1

The given relationship is x3 + x2y + xy2 + y3 = 81
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(x3 + x2y + xy2 + y3) = $\frac {d}{dx}$(81)

⇒ $\frac {d}{dx}$(x3) + $\frac {d}{dx}$(x2y) + $\frac {d}{dx}$(xy2) + $\frac {d}{dx}$(y3)=0

⇒ 3x2 + [y.$\frac {d}{dx}$(x2) + x2.$\frac {dy}{dx}$] + [y2 $\frac {d}{dx}$(x) +x $\frac {d}{dx}$(y2)] + 3y2$\frac {d}{dx}$ = 0

⇒ 3x2 + [y.2x + x2$\frac {dy}{dx}$] + [y2.1 + x.2y.$\frac {dy}{dx}$] + 3y2$\frac {dy}{dx}$ = 0

⇒ (x2 + 2xy + 3y2)$\frac {dy}{dx}$ + (3x2 + 2xy + y2) = 0

∴ $\frac {dy}{dx}$ = $-\frac {(3x^2+2xy+y^2)}{(x^2+2xy+3y^2)}$