Question

question: find $\frac{dy}{dx}$, where $y$ is defined as a function of $x$ implicitly by the equation below. $y^{5}-2x^{4}y^{2}=1$. select the correct answer below: $\frac{dy}{dx}=\frac{8x^{3}y^{2}}{5y^{4}-4x^{4}y}$, $\frac{dy}{dx}=\frac{8x^{3}y^{2}-4x^{4}y}{5y^{4}}$, $\frac{dy}{dx}=\frac{8x^{3}y^{2}+4x^{4}y}{5y^{4}}$, $\frac{dy}{dx}=\frac{8x^{3}y^{2}}{5y^{4}+4x^{4}y}$

Explanation:

Step1: Differentiate both sides

Differentiate $y^{5}-2x^{4}y^{2}=1$ with respect to $x$ using product - rule and chain - rule.
$5y^{4}\frac{dy}{dx}-(8x^{3}y^{2}+4x^{4}y\frac{dy}{dx}) = 0$

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

$5y^{4}\frac{dy}{dx}-4x^{4}y\frac{dy}{dx}=8x^{3}y^{2}$
$\frac{dy}{dx}(5y^{4}-4x^{4}y)=8x^{3}y^{2}$
$\frac{dy}{dx}=\frac{8x^{3}y^{2}}{5y^{4}-4x^{4}y}$

Answer:

$\frac{dy}{dx}=\frac{8x^{3}y^{2}}{5y^{4}-4x^{4}y}$