Question

question
find the differential (dy) of the function
(y=\frac{-2x^{3}-6}{2x - 3})
provide your answer below:
(dy=square dx)

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Here, $u = 2x - 3$ and $v=-2x^{3}-6$.

Step2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$

Differentiate $u = 2x - 3$ with respect to $x$: $\frac{du}{dx}=2$. Differentiate $v=-2x^{3}-6$ with respect to $x$: $\frac{dv}{dx}=-6x^{2}$.

Step3: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{(-2x^{3}-6)\times2-(2x - 3)\times(-6x^{2})}{(-2x^{3}-6)^{2}}$.
Expand the numerator:
\[

$$\begin{align*} &(-2x^{3}-6)\times2-(2x - 3)\times(-6x^{2})\\ =&-4x^{3}-12+(12x^{3}-18x^{2})\\ =&-4x^{3}-12 + 12x^{3}-18x^{2}\\ =&8x^{3}-18x^{2}-12 \end{align*}$$

\]
So, $\frac{dy}{dx}=\frac{8x^{3}-18x^{2}-12}{(-2x^{3}-6)^{2}}$.

Answer:

$\frac{8x^{3}-18x^{2}-12}{(-2x^{3}-6)^{2}}$