Question

If (x² + y²)² = xy, then `dy/dx` is ______.

Options

• `(y + 4x(x^2 + y^2))/(4y(x^2 + y^2) - x)`

• `(y - 4x(x^2 + y^2))/(x + 4(x^2 + y^2))`

• `(y - 4x(x^2 + y^2))/(4y(x^2 + y^2) - x)`

• `(4y(x^2 + y^2) - x)/(y - 4x(x^2 + y^2))`

Answer 1

If (x² + y²)² = xy, then `dy/dx` is `underlinebb((y - 4x(x^2 + y^2))/(4y(x^2 + y^2) - x))`.

Explanation:

Given, (x² + y²)² = xy

`\implies` x⁴ + 2x²y² + y⁴ – xy = 0

Differentiating w.r.t. x, we get

`4x^3 + 2[2xy^2 + x^2 . 2y dy/dx] + 4y^3 dy/dx - [y + x dy/dx] = 0`

`dy/dx [4x^2y + 4y^3 - x] + [4x^3 + 4xy^2 - y] = 0`

`dy/dx = (-[4x^3 + 4xy^2 - y])/([4x^2y + 4y^3 - x])`

or `dy/dx = (y - 4x(x^2 + y^2))/(4y(x^2 + y^2) - x)`