Answer to Question #269321 in Calculus for Angel Nodado

Question

Find dy.

xy^4 + x ^ 2 y = x + 3y

Answer 1

$xy^4 +x^2y = x +3y\\ \text{Differentiating implicitly, we have that}\\ y^4 + 4xy^3\frac{dy}{dx} +2xy + x^2 \frac{dy}{dx}= 1 +3\frac{dy}{dx}\\ \text{Collecting like terms, we have that}\\ \frac{dy}{dx}(4xy^3+x^2-3)= 1-2xy-y^4\\ \implies \frac{dy}{dx} = \frac{1-2xy-y^4}{4xy^3+x^2-3}$