Question

question 24 - 1 point. find \\(\frac{dy}{dx}\\), where \\(y\\) is defined as a function of \\(x\\) implicitly by the equation below. \\(y^{3}+2x^{2}y^{5}=1\\). select the correct answer below. \\(\frac{dy}{dx}=\frac{- 10x^{2}y^{4}-3y^{2}}{4xy^{5}}\\), \\(\frac{dy}{dx}=\frac{10x^{2}y^{4}-3y^{2}}{4xy^{5}}\\), \\(\frac{dy}{dx}=\frac{10x^{2}y^{4}+3y^{2}}{4xy^{5}}\\), \\(\frac{dy}{dx}=\frac{- 10x^{2}y^{4}+3y^{2}}{4xy^{5}}\\)

Explanation:

Step1: Differentiate both sides

Differentiate $y^{3}+2x^{2}y^{5}=1$ with respect to $x$. Using product - rule and chain - rule, we get $3y^{2}\frac{dy}{dx}+4xy^{5}+10x^{2}y^{4}\frac{dy}{dx}=0$.

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

Isolate $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(3y^{2}+10x^{2}y^{4})=- 4xy^{5}$. Then $\frac{dy}{dx}=-\frac{4xy^{5}}{10x^{2}y^{4}+3y^{2}}$.

Answer:

$\frac{dy}{dx}=-\frac{4xy^{5}}{10x^{2}y^{4}+3y^{2}}$