Question

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

Explanation:

Step1: Differentiate both sides

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

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

Factor out $\frac{dy}{dx}$ on the right - hand side: $-2x + 3=(4y - 2)\frac{dy}{dx}$. Then $\frac{dy}{dx}=\frac{-2x + 3}{4y-2}$.

Answer:

$\frac{dy}{dx}=\frac{-2x + 3}{4y-2}$