If $x^2+2xy +y^3=10,$ then value of $\fr

CUET Preparation Today

Start Free Mocks

Home Questions QNPQ4T8VKG

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

If $x^2+2xy +y^3=10,$ then value of $\frac{dy}{dx}$ is :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

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

$x^2+2xy +y^3=10$

differentiating eq. wrt (x)

$2x+2y+2x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$

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