Question

question. find $\frac{dy}{dx}$, where $y$ is defined as a function of $x$ implicitly by the equation below. $y - 3x^{2}y^{5}=-2$. select the correct answer below: $\frac{dy}{dx}=\frac{6xy^{5}}{15x^{2}y^{4}-1}$, $\frac{dy}{dx}=\frac{6xy^{5}}{15x^{2}y^{4}+1}$, $\frac{dy}{dx}=\frac{6xy^{5}}{-15x^{2}y^{4}-1}$, $\frac{dy}{dx}=\frac{6xy^{5}}{-15x^{2}y^{4}+1}$

Explanation:

Step1: Differentiate both sides

Differentiate $y - 3x^{2}y^{5}=-2$ with respect to $x$. Using product - rule and chain - rule, we get $\frac{dy}{dx}-(6xy^{5}+15x^{2}y^{4}\frac{dy}{dx}) = 0$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange terms: $\frac{dy}{dx}-6xy^{5}-15x^{2}y^{4}\frac{dy}{dx}=0$. Then $\frac{dy}{dx}(1 - 15x^{2}y^{4})=6xy^{5}$. So $\frac{dy}{dx}=\frac{6xy^{5}}{1 - 15x^{2}y^{4}}$.

Answer:

$\frac{dy}{dx}=\frac{6xy^{5}}{1 - 15x^{2}y^{4}}$