If $x^3+y^3=xy,$ then $\frac{dy}{dx}$ is

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Home Questions PBSQ9S635A

Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Calculus

Question:

If $x^3+y^3=xy,$ then $\frac{dy}{dx}$ is equal to :

Options:

$\frac{y+3x^2}{3y^2+x}$

$\frac{y-3x^2}{3y^2-x}$

$\frac{y+3x^2}{3y^2-x}$

$\frac{2y-x^2}{y^2-3x}$

Correct Answer:

$\frac{y-3x^2}{3y^2-x}$

Explanation:

The correct answer is Option (2) → $\frac{y-3x^2}{3y^2-x}$

$x^3+y^3=xy$

Differentiate w.r.t 'x'

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

$⇒\frac{dy}{dx}(3y^2-x)=y-3x^2$

$⇒\frac{dy}{dx}=\frac{y-3x^2}{3y^2-x}$