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Identifying \(\displaystyle f=- 8 x - 7\) and \(\displaystyle g=\left(- 8 x - 8\right) \left(3 x + 1\right)\) and using the product rule with \(\displaystyle f=- 8 x - 7 \implies f'=-8\). This leaves g as \(\displaystyle g = \left(- 8 x - 8\right) \left(3 x + 1\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=3 x + 1 \implies f'=3\) and \(\displaystyle g=- 8 x - 8 \implies g'=-8\). Popping up a level gives \(\displaystyle g'=(- 8 x - 8)(3)+(3 x + 1)(-8)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- 8 x - 7)(- 48 x - 32)+(\left(- 8 x - 8\right) \left(3 x + 1\right))(-8)=\left(- 24 x - 24\right) \left(- 8 x - 7\right) + \left(- 24 x - 8\right) \left(- 8 x - 7\right) - 8 \left(- 8 x - 8\right) \left(3 x + 1\right)\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (- 8 x - 7)(3 x + 1)(- 8 x - 8)$.
    \soln{9cm}{Identifying $f=- 8 x - 7$ and $g=\left(- 8 x - 8\right) \left(3 x + 1\right)$ and using the product rule with $f=- 8 x - 7 \implies f'=-8$. This leaves g as $g = \left(- 8 x - 8\right) \left(3 x + 1\right)$ which also requires the product rule. Pushing down in the new product rule $f=3 x + 1 \implies f'=3$ and $g=- 8 x - 8 \implies g'=-8$. Popping up a level gives $g'=(- 8 x - 8)(3)+(3 x + 1)(-8)$Popping up again (Back to the original problem) gives $f'=(- 8 x - 7)(- 48 x - 32)+(\left(- 8 x - 8\right) \left(3 x + 1\right))(-8)=\left(- 24 x - 24\right) \left(- 8 x - 7\right) + \left(- 24 x - 8\right) \left(- 8 x - 7\right) - 8 \left(- 8 x - 8\right) \left(3 x + 1\right)$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (- 8 x - 7)(3 x + 1)(- 8 x - 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28-%208%20x%20-%207%29%283%20x%20%2B%201%29%28-%208%20x%20-%208%29%20" alt="LaTeX:  \displaystyle y = (- 8 x - 7)(3 x + 1)(- 8 x - 8) " data-equation-content=" \displaystyle y = (- 8 x - 7)(3 x + 1)(- 8 x - 8) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=- 8 x - 7 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%20-%207%20" alt="LaTeX:  \displaystyle f=- 8 x - 7 " data-equation-content=" \displaystyle f=- 8 x - 7 " />  and  <img class="equation_image" title=" \displaystyle g=\left(- 8 x - 8\right) \left(3 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%208%20x%20-%208%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(- 8 x - 8\right) \left(3 x + 1\right) " data-equation-content=" \displaystyle g=\left(- 8 x - 8\right) \left(3 x + 1\right) " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=- 8 x - 7 \implies f'=-8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%20-%207%20%5Cimplies%20f%27%3D-8%20" alt="LaTeX:  \displaystyle f=- 8 x - 7 \implies f'=-8 " data-equation-content=" \displaystyle f=- 8 x - 7 \implies f'=-8 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(- 8 x - 8\right) \left(3 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%208%20x%20-%208%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(- 8 x - 8\right) \left(3 x + 1\right) " data-equation-content=" \displaystyle g = \left(- 8 x - 8\right) \left(3 x + 1\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=3 x + 1 \implies f'=3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D3%20x%20%2B%201%20%5Cimplies%20f%27%3D3%20" alt="LaTeX:  \displaystyle f=3 x + 1 \implies f'=3 " data-equation-content=" \displaystyle f=3 x + 1 \implies f'=3 " />  and  <img class="equation_image" title=" \displaystyle g=- 8 x - 8 \implies g'=-8 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%208%20x%20-%208%20%5Cimplies%20g%27%3D-8%20" alt="LaTeX:  \displaystyle g=- 8 x - 8 \implies g'=-8 " data-equation-content=" \displaystyle g=- 8 x - 8 \implies g'=-8 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(- 8 x - 8)(3)+(3 x + 1)(-8) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%208%20x%20-%208%29%283%29%2B%283%20x%20%2B%201%29%28-8%29%20" alt="LaTeX:  \displaystyle g'=(- 8 x - 8)(3)+(3 x + 1)(-8) " data-equation-content=" \displaystyle g'=(- 8 x - 8)(3)+(3 x + 1)(-8) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(- 8 x - 7)(- 48 x - 32)+(\left(- 8 x - 8\right) \left(3 x + 1\right))(-8)=\left(- 24 x - 24\right) \left(- 8 x - 7\right) + \left(- 24 x - 8\right) \left(- 8 x - 7\right) - 8 \left(- 8 x - 8\right) \left(3 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%208%20x%20-%207%29%28-%2048%20x%20-%2032%29%2B%28%5Cleft%28-%208%20x%20-%208%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%29%28-8%29%3D%5Cleft%28-%2024%20x%20-%2024%5Cright%29%20%5Cleft%28-%208%20x%20-%207%5Cright%29%20%2B%20%5Cleft%28-%2024%20x%20-%208%5Cright%29%20%5Cleft%28-%208%20x%20-%207%5Cright%29%20-%208%20%5Cleft%28-%208%20x%20-%208%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle f'=(- 8 x - 7)(- 48 x - 32)+(\left(- 8 x - 8\right) \left(3 x + 1\right))(-8)=\left(- 24 x - 24\right) \left(- 8 x - 7\right) + \left(- 24 x - 8\right) \left(- 8 x - 7\right) - 8 \left(- 8 x - 8\right) \left(3 x + 1\right) " data-equation-content=" \displaystyle f'=(- 8 x - 7)(- 48 x - 32)+(\left(- 8 x - 8\right) \left(3 x + 1\right))(-8)=\left(- 24 x - 24\right) \left(- 8 x - 7\right) + \left(- 24 x - 8\right) \left(- 8 x - 7\right) - 8 \left(- 8 x - 8\right) \left(3 x + 1\right) " /> </p> </p>