question 17 · 1 point for k(x)=(x - 1)(-x² - 2x)(2x - 2), find the derivative of k(x) using the product rule. do not include \k(x)=\ in your answer. for example, if your answer is k(x)=4x⁴, you would enter 4x⁴. provide your answer below:

Question

Accepted Answer

# Explanation:
## Step1: Recall product - rule
The product rule states that if $y = uvw$, then $y'=u'vw + uv'w+uvw'$. Let $u=(x - 1)$, $v=(-x^{2}-2x)$, $w=(2x - 2)$.
## Step2: Find derivatives of $u$, $v$, $w$
$u' = 1$, $v'=-2x - 2$, $w'=2$.
## Step3: Apply product - rule
$$
\begin{align*}
&(1)\times(-x^{2}-2x)\times(2x - 2)+(x - 1)\times(-2x - 2)\times(2x - 2)+(x - 1)\times(-x^{2}-2x)\times2\
=&(-x^{2}-2x)(2x - 2)+(x - 1)(-2x - 2)(2x - 2)+2(x - 1)(-x^{2}-2x)
\end{align*}
$$
Expand and simplify:
$$
\begin{align*}
(-x^{2}-2x)(2x - 2)&=-2x^{3}+2x^{2}-4x^{2}+4x=-2x^{3}-2x^{2}+4x\
(x - 1)(-2x - 2)(2x - 2)&=(x - 1)[(-2x - 2)(2x - 2)]=(x - 1)(-4x^{2}+4x + 4x - 4)=(x - 1)(-4x^{2}+8x - 4)\
&=-4x^{3}+8x^{2}-4x + 4x^{2}-8x + 4=-4x^{3}+12x^{2}-12x + 4\
2(x - 1)(-x^{2}-2x)&=2(-x^{3}-2x^{2}+x^{2}+2x)=-2x^{3}-2x^{2}+4x
\end{align*}
$$
Sum them up:
$$
\begin{align*}
&(-2x^{3}-2x^{2}+4x)+(-4x^{3}+12x^{2}-12x + 4)+(-2x^{3}-2x^{2}+4x)\
=&(-2x^{3}-4x^{3}-2x^{3})+(-2x^{2}+12x^{2}-2x^{2})+(4x-12x + 4x)+4\
=&-8x^{3}+8x^{2}-4x + 4
\end{align*}
$$

# Answer:
$-8x^{3}+8x^{2}-4x + 4$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall product - rule

Step2: Find derivatives of \(u\), \(v\), \(w\)

Step3: Apply product - rule

Answer: