Question

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

Explanation:

Step1: Rewrite the function

Rewrite $y=\frac{4 - 3x^{3}}{5-5x}$ as a quotient of two functions $u = 4 - 3x^{3}$ and $v=5 - 5x$.

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

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $u'=-9x^{2}$ and $v'=-5$.

Step3: Apply the quotient rule

The quotient rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Substitute $u$, $u'$, $v$, and $v'$ into the quotient rule:
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{(-9x^{2})(5 - 5x)-(4 - 3x^{3})(-5)}{(5 - 5x)^{2}}\\ &=\frac{-45x^{2}+45x^{3}+20 - 15x^{3}}{(5 - 5x)^{2}}\\ &=\frac{30x^{3}-45x^{2}+20}{(5 - 5x)^{2}} \end{align*}$$

\]

Answer:

$\frac{30x^{3}-45x^{2}+20}{(5 - 5x)^{2}}$