Answer
For $y^5+x^2y^3=1+x^4y$ find $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{4x^3y-2xy^3}{5y^4+3x^2y^2-x^4}$
Work Step by Step
Differentiate both sides with respect to x
$5y^4\frac{dy}{dx}+2xy^3+3x^2y^2\frac{dy}{dx}=4x^3y+x^4\frac{dy}{dx}$
Isolate $\frac{dy}{dx}$
$5y^4\frac{dy}{dx}+3x^2y^2\frac{dy}{dx}-x^4\frac{dy}{dx}=4x^3y-2xy^3$
$\frac{dy}{dx}(5y^4+3x^2y^2-x^4)=4x^3y-2xy^3$
$\frac{dy}{dx}=\frac{4x^3y-2xy^3}{5y^4+3x^2y^2-x^4}$
