Question

if `(x^2 + y^2)^2 = xy` find `(dy)/(dx)`

Answer 1

We have `(x^2 + y^2) = xy`

Differentiating with respect to x, we get

`=> d/dx [(x^2 + y^2)^2] = d/dx (xy)`

`=> 2(x^2 + y^2) d/(dx) (x^2 + y^2) = x (dy)/(dx) + y d/dx (x)`

`=> 2(x^2 + y^2) (2x+ 2y dy/dx) = x (dy/dx) + y (1)`

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

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

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

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

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