Question

question. find the differential $dy$ of the function $y=\frac{-4x^{3}-10}{5x - 2}$. provide your answer below. $\frac{dy}{dx}=square$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$, where $u=-4x^{3}-10$ and $v = 5x - 2$.

Step2: Find $u'$ and $v'$

Differentiate $u=-4x^{3}-10$ with respect to $x$: $u'=-12x^{2}$. Differentiate $v = 5x - 2$ with respect to $x$: $v'=5$.

Step3: Apply the quotient - rule

$\frac{dy}{dx}=\frac{(-12x^{2})(5x - 2)-(-4x^{3}-10)\times5}{(5x - 2)^{2}}$.
Expand the numerator:
\[

$$\begin{align*} (-12x^{2})(5x - 2)-(-4x^{3}-10)\times5&=-60x^{3}+24x^{2}+20x^{3}+50\\ &=(-60x^{3}+20x^{3})+24x^{2}+50\\ &=-40x^{3}+24x^{2}+50 \end{align*}$$

\]
So, $\frac{dy}{dx}=\frac{-40x^{3}+24x^{2}+50}{(5x - 2)^{2}}$.

Answer:

$\frac{-40x^{3}+24x^{2}+50}{(5x - 2)^{2}}$