## Calculus 8th Edition

For $x^4+x^2y^2+y^3=5$ find $\frac{dy}{dx}$
Differentiate both sides with respect to x $4x^3+2xy^2+x^22y\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$ Isolate $\frac{dy}{dx}$ $\frac{dy}{dx}(x^22y+3y^2)=-4x^3-2xy^2$ $\frac{dy}{dx}=\frac{-4x^3-2xy^2}{x^22y+3y^2}$