Question

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

Explanation:

Step1: Differentiate both sides

Differentiate $-4x^{2}-3x - 1=y^{2}-y$ with respect to $x$. Using the power - rule, we get $-8x - 3 = 2y\frac{dy}{dx}-\frac{dy}{dx}$.

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

Factor out $\frac{dy}{dx}$ on the right - hand side: $-8x - 3=(2y - 1)\frac{dy}{dx}$. Then $\frac{dy}{dx}=\frac{-8x - 3}{2y - 1}$.

Answer:

$\frac{dy}{dx}=\frac{-8x - 3}{2y - 1}$