Question

find dy/dx of the following: x^3+xy^2=y^3+yx^2

Answer 1

ANSWER:

• dy/dx = (3x² + y² - 2xy) / (3y² + x² - 2xy).

GIVEN:

• x³ + xy² = y³ + yx².

TO FIND:

• The value of dy/dx.

EXPLANATION:

$\implies \sf x^3+xy^2=y^3+yx^2 \\ \\ \\ \mapsto\sf Differentiate\ w.r.t\ x.\\ \\ \\ \bigstar \ \boxed{ \bold{ \large{ \pink{ \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }}}} \\ \\ \\ \: \bigstar \ \boxed{ \bold{ \large{ \blue{ \dfrac{d}{dx} uv =uv' + vu' }}}} \\ \\ \\ \implies \sf 3x^2 \dfrac{dx}{dx} + y^2\dfrac{dx}{dx}+2xy\dfrac{dy}{dx}=3y^2\dfrac{dy}{dx} + \dfrac{dy}{dx}x^2+2xy\dfrac{dx}{dx} \\ \\ \\ \implies \sf 3x^2+y^2+2xy\dfrac{dy}{dx}=3y^2\dfrac{dy}{dx} +x^2 \dfrac{dy}{dx}+2xy \\ \\ \\ \implies \sf 3x^2 + y^2 - 2xy= \dfrac{dy}{dx} (3 {y}^{2} +x^2 - 2xy ) \\ \\ \\ \sf \implies\dfrac{dy}{dx} = \dfrac{3x^2+y^2 - 2xy}{3 {y}^{2}+x^2 - 2xy}$