Question

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:

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:

\(-8x^{3}+8x^{2}-4x + 4\)