Question

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

Explanation:

Step1: Differentiate both sides

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

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

Isolate $\frac{dy}{dx}$:
\[

$$\begin{align*} 2x^{3}y\frac{dy}{dx}&=- 5x^{4}-3x^{2}y^{2}\\ \frac{dy}{dx}&=\frac{-5x^{4}-3x^{2}y^{2}}{2x^{3}y}\\ \frac{dy}{dx}&=-\frac{5x}{2y}-\frac{3x}{2y}\\ \end{align*}$$

\]

Answer:

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