Find dx/dy for (x^2+y^2)^2= 2x^2*y?

Question

Note that the problem is asking for dx/dy and NOT dy/dx. Y is the independent variable.

Answer 1

I apologize, I read the previous question too quickly, and 99.99999999% of these I've ever seen ask for dy/dx, not dx/dy.

Nonetheless, the approach is the same.

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

Expand:

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

Rearrange so dx/dy terms are on one side:

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

Factor out dx/dy:

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

Solve for dx/dy:

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