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Identifying \(\displaystyle f=- 8 x^{2} + 3 x + 8\) and \(\displaystyle g=\left(- 6 x^{2} + 4 x + 8\right) e^{x}\) and using the product rule with \(\displaystyle f=- 8 x^{2} + 3 x + 8 \implies f'=3 - 16 x\). This leaves g as \(\displaystyle g = \left(- 6 x^{2} + 4 x + 8\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=e^{x} \implies f'=e^{x}\) and \(\displaystyle g=- 6 x^{2} + 4 x + 8 \implies g'=4 - 12 x\). Popping up a level gives \(\displaystyle g'=(- 6 x^{2} + 4 x + 8)(e^{x})+(e^{x})(4 - 12 x)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- 8 x^{2} + 3 x + 8)(\left(4 - 12 x\right) e^{x} + \left(- 6 x^{2} + 4 x + 8\right) e^{x})+(\left(- 6 x^{2} + 4 x + 8\right) e^{x})(3 - 16 x)=\left(3 - 16 x\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} + \left(4 - 12 x\right) \left(- 8 x^{2} + 3 x + 8\right) e^{x} + \left(- 8 x^{2} + 3 x + 8\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x}\)

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\begin{question}Find the derivative of $y = (- 8 x^{2} + 3 x + 8)(e^{x})(- 6 x^{2} + 4 x + 8)$.
    \soln{9cm}{Identifying $f=- 8 x^{2} + 3 x + 8$ and $g=\left(- 6 x^{2} + 4 x + 8\right) e^{x}$ and using the product rule with $f=- 8 x^{2} + 3 x + 8 \implies f'=3 - 16 x$. This leaves g as $g = \left(- 6 x^{2} + 4 x + 8\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=e^{x} \implies f'=e^{x}$ and $g=- 6 x^{2} + 4 x + 8 \implies g'=4 - 12 x$. Popping up a level gives $g'=(- 6 x^{2} + 4 x + 8)(e^{x})+(e^{x})(4 - 12 x)$Popping up again (Back to the original problem) gives $f'=(- 8 x^{2} + 3 x + 8)(\left(4 - 12 x\right) e^{x} + \left(- 6 x^{2} + 4 x + 8\right) e^{x})+(\left(- 6 x^{2} + 4 x + 8\right) e^{x})(3 - 16 x)=\left(3 - 16 x\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} + \left(4 - 12 x\right) \left(- 8 x^{2} + 3 x + 8\right) e^{x} + \left(- 8 x^{2} + 3 x + 8\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x}$}

\end{question}

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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^{2} + 3 x + 8)(e^{x})(- 6 x^{2} + 4 x + 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%29%28e%5E%7Bx%7D%29%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%29%20" alt="LaTeX:  \displaystyle y = (- 8 x^{2} + 3 x + 8)(e^{x})(- 6 x^{2} + 4 x + 8) " data-equation-content=" \displaystyle y = (- 8 x^{2} + 3 x + 8)(e^{x})(- 6 x^{2} + 4 x + 8) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=- 8 x^{2} + 3 x + 8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%20" alt="LaTeX:  \displaystyle f=- 8 x^{2} + 3 x + 8 " data-equation-content=" \displaystyle f=- 8 x^{2} + 3 x + 8 " />  and  <img class="equation_image" title=" \displaystyle g=\left(- 6 x^{2} + 4 x + 8\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(- 6 x^{2} + 4 x + 8\right) e^{x} " data-equation-content=" \displaystyle g=\left(- 6 x^{2} + 4 x + 8\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=- 8 x^{2} + 3 x + 8 \implies f'=3 - 16 x " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%20%5Cimplies%20f%27%3D3%20-%2016%20x%20" alt="LaTeX:  \displaystyle f=- 8 x^{2} + 3 x + 8 \implies f'=3 - 16 x " data-equation-content=" \displaystyle f=- 8 x^{2} + 3 x + 8 \implies f'=3 - 16 x " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(- 6 x^{2} + 4 x + 8\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(- 6 x^{2} + 4 x + 8\right) e^{x} " data-equation-content=" \displaystyle g = \left(- 6 x^{2} + 4 x + 8\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=e^{x} \implies f'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20%5Cimplies%20f%27%3De%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f=e^{x} \implies f'=e^{x} " data-equation-content=" \displaystyle f=e^{x} \implies f'=e^{x} " />  and  <img class="equation_image" title=" \displaystyle g=- 6 x^{2} + 4 x + 8 \implies g'=4 - 12 x " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%20%5Cimplies%20g%27%3D4%20-%2012%20x%20" alt="LaTeX:  \displaystyle g=- 6 x^{2} + 4 x + 8 \implies g'=4 - 12 x " data-equation-content=" \displaystyle g=- 6 x^{2} + 4 x + 8 \implies g'=4 - 12 x " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(- 6 x^{2} + 4 x + 8)(e^{x})+(e^{x})(4 - 12 x) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%29%28e%5E%7Bx%7D%29%2B%28e%5E%7Bx%7D%29%284%20-%2012%20x%29%20" alt="LaTeX:  \displaystyle g'=(- 6 x^{2} + 4 x + 8)(e^{x})+(e^{x})(4 - 12 x) " data-equation-content=" \displaystyle g'=(- 6 x^{2} + 4 x + 8)(e^{x})+(e^{x})(4 - 12 x) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(- 8 x^{2} + 3 x + 8)(\left(4 - 12 x\right) e^{x} + \left(- 6 x^{2} + 4 x + 8\right) e^{x})+(\left(- 6 x^{2} + 4 x + 8\right) e^{x})(3 - 16 x)=\left(3 - 16 x\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} + \left(4 - 12 x\right) \left(- 8 x^{2} + 3 x + 8\right) e^{x} + \left(- 8 x^{2} + 3 x + 8\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%29%28%5Cleft%284%20-%2012%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%29%283%20-%2016%20x%29%3D%5Cleft%283%20-%2016%20x%5Cright%29%20%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%284%20-%2012%20x%5Cright%29%20%5Cleft%28-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%208%20x%5E%7B2%7D%20%2B%203%20x%20%2B%208%5Cright%29%20%5Cleft%28-%206%20x%5E%7B2%7D%20%2B%204%20x%20%2B%208%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(- 8 x^{2} + 3 x + 8)(\left(4 - 12 x\right) e^{x} + \left(- 6 x^{2} + 4 x + 8\right) e^{x})+(\left(- 6 x^{2} + 4 x + 8\right) e^{x})(3 - 16 x)=\left(3 - 16 x\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} + \left(4 - 12 x\right) \left(- 8 x^{2} + 3 x + 8\right) e^{x} + \left(- 8 x^{2} + 3 x + 8\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} " data-equation-content=" \displaystyle f'=(- 8 x^{2} + 3 x + 8)(\left(4 - 12 x\right) e^{x} + \left(- 6 x^{2} + 4 x + 8\right) e^{x})+(\left(- 6 x^{2} + 4 x + 8\right) e^{x})(3 - 16 x)=\left(3 - 16 x\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} + \left(4 - 12 x\right) \left(- 8 x^{2} + 3 x + 8\right) e^{x} + \left(- 8 x^{2} + 3 x + 8\right) \left(- 6 x^{2} + 4 x + 8\right) e^{x} " /> </p> </p>