Question

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

Explanation:

Step1: Simplify the function

First, simplify $y=\frac{-2x^{3}-6}{2x - 3}$. We can use the quotient - rule for differentiation. The quotient rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=-2x^{3}-6$ and $v = 2x - 3$.
$u^\prime=-6x^{2}$ and $v^\prime = 2$.

Step2: Apply the quotient - rule

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

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

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

Answer:

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