Question

Find $\frac {dy}{dx}$: $x^2+xy+y^2=100$

Answer 1

The given relationship is x2 + xy + y2 = 100
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(x2 + xy + y2) = $\frac {d}{dx}$(100)

⇒ $\frac {d}{dx}$(x2) + $\frac {d}{dx}$(xy) + $\frac {d}{dx}$(y2)=0

⇒ 2x + [y . $\frac {d}{dx}$(x) + x . $\frac {dy}{dx}$] + 2y $\frac {dy}{dx}$ = 0 [using product rule and chain rule]

⇒ 2x + y . 1 + x . $\frac {dy}{dx}$ + 2y $\frac {dy}{dx}$ = 0

⇒ 2x + y + (x+2y) $\frac {dy}{dx}$ = 0

∴ $\frac {dy}{dx}$ = $-\frac {2x+y}{x+2y}$