Question

Find $\frac{dy}{dx},$ where $y$ is defined as a function of $x$ implicitly by the equation below.

$y^2-3x^3{y}^4=1$

Select the correct answer below:

$\frac{dy}{dx}=\frac{9x^2{y}^4}{-12x^3{y}^{3}2y}$

$\frac{dy}{dx}=\frac{9x^2{y}^4}{12x^3{y}^3-2y}$

$\frac{dy}{dx}=\frac{9x^2{y}^4}{-12x^3{y}^3-2y}$

$\frac{dy}{dx}=\frac{9x^2{y}^4}{12x^3{y}^{3}2y}$