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Identifying \(\displaystyle f=x^{2} - 2 x + 6\) and \(\displaystyle g=\left(- 2 x^{2} - 2 x + 1\right) e^{x}\) and using the product rule with \(\displaystyle f=x^{2} - 2 x + 6 \implies f'=2 x - 2\). This leaves g as \(\displaystyle g = \left(- 2 x^{2} - 2 x + 1\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 2 x^{2} - 2 x + 1 \implies f'=- 4 x - 2\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(- 4 x - 2)+(- 2 x^{2} - 2 x + 1)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(x^{2} - 2 x + 6)(\left(- 4 x - 2\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) e^{x})+(\left(- 2 x^{2} - 2 x + 1\right) e^{x})(2 x - 2)=\left(- 4 x - 2\right) \left(x^{2} - 2 x + 6\right) e^{x} + \left(2 x - 2\right) \left(- 2 x^{2} - 2 x + 1\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) \left(x^{2} - 2 x + 6\right) e^{x}\)

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\begin{question}Find the derivative of $y = (x^{2} - 2 x + 6)(- 2 x^{2} - 2 x + 1)(e^{x})$.
    \soln{9cm}{Identifying $f=x^{2} - 2 x + 6$ and $g=\left(- 2 x^{2} - 2 x + 1\right) e^{x}$ and using the product rule with $f=x^{2} - 2 x + 6 \implies f'=2 x - 2$. This leaves g as $g = \left(- 2 x^{2} - 2 x + 1\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=- 2 x^{2} - 2 x + 1 \implies f'=- 4 x - 2$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(- 4 x - 2)+(- 2 x^{2} - 2 x + 1)(e^{x})$Popping up again (Back to the original problem) gives $f'=(x^{2} - 2 x + 6)(\left(- 4 x - 2\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) e^{x})+(\left(- 2 x^{2} - 2 x + 1\right) e^{x})(2 x - 2)=\left(- 4 x - 2\right) \left(x^{2} - 2 x + 6\right) e^{x} + \left(2 x - 2\right) \left(- 2 x^{2} - 2 x + 1\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) \left(x^{2} - 2 x + 6\right) e^{x}$}

\end{question}

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

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HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (x^{2} - 2 x + 6)(- 2 x^{2} - 2 x + 1)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28x%5E%7B2%7D%20-%202%20x%20%2B%206%29%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle y = (x^{2} - 2 x + 6)(- 2 x^{2} - 2 x + 1)(e^{x}) " data-equation-content=" \displaystyle y = (x^{2} - 2 x + 6)(- 2 x^{2} - 2 x + 1)(e^{x}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=x^{2} - 2 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B2%7D%20-%202%20x%20%2B%206%20" alt="LaTeX:  \displaystyle f=x^{2} - 2 x + 6 " data-equation-content=" \displaystyle f=x^{2} - 2 x + 6 " />  and  <img class="equation_image" title=" \displaystyle g=\left(- 2 x^{2} - 2 x + 1\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(- 2 x^{2} - 2 x + 1\right) e^{x} " data-equation-content=" \displaystyle g=\left(- 2 x^{2} - 2 x + 1\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=x^{2} - 2 x + 6 \implies f'=2 x - 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B2%7D%20-%202%20x%20%2B%206%20%5Cimplies%20f%27%3D2%20x%20-%202%20" alt="LaTeX:  \displaystyle f=x^{2} - 2 x + 6 \implies f'=2 x - 2 " data-equation-content=" \displaystyle f=x^{2} - 2 x + 6 \implies f'=2 x - 2 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(- 2 x^{2} - 2 x + 1\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(- 2 x^{2} - 2 x + 1\right) e^{x} " data-equation-content=" \displaystyle g = \left(- 2 x^{2} - 2 x + 1\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=- 2 x^{2} - 2 x + 1 \implies f'=- 4 x - 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%20%5Cimplies%20f%27%3D-%204%20x%20-%202%20" alt="LaTeX:  \displaystyle f=- 2 x^{2} - 2 x + 1 \implies f'=- 4 x - 2 " data-equation-content=" \displaystyle f=- 2 x^{2} - 2 x + 1 \implies f'=- 4 x - 2 " />  and  <img class="equation_image" title=" \displaystyle g=e^{x} \implies g'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3De%5E%7Bx%7D%20%5Cimplies%20g%27%3De%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=e^{x} \implies g'=e^{x} " data-equation-content=" \displaystyle g=e^{x} \implies g'=e^{x} " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(e^{x})(- 4 x - 2)+(- 2 x^{2} - 2 x + 1)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%28-%204%20x%20-%202%29%2B%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle g'=(e^{x})(- 4 x - 2)+(- 2 x^{2} - 2 x + 1)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(- 4 x - 2)+(- 2 x^{2} - 2 x + 1)(e^{x}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(x^{2} - 2 x + 6)(\left(- 4 x - 2\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) e^{x})+(\left(- 2 x^{2} - 2 x + 1\right) e^{x})(2 x - 2)=\left(- 4 x - 2\right) \left(x^{2} - 2 x + 6\right) e^{x} + \left(2 x - 2\right) \left(- 2 x^{2} - 2 x + 1\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) \left(x^{2} - 2 x + 6\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28x%5E%7B2%7D%20-%202%20x%20%2B%206%29%28%5Cleft%28-%204%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%29%282%20x%20-%202%29%3D%5Cleft%28-%204%20x%20-%202%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%202%20x%20%2B%206%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%282%20x%20-%202%5Cright%29%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%202%20x%20%2B%201%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%202%20x%20%2B%206%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(x^{2} - 2 x + 6)(\left(- 4 x - 2\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) e^{x})+(\left(- 2 x^{2} - 2 x + 1\right) e^{x})(2 x - 2)=\left(- 4 x - 2\right) \left(x^{2} - 2 x + 6\right) e^{x} + \left(2 x - 2\right) \left(- 2 x^{2} - 2 x + 1\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) \left(x^{2} - 2 x + 6\right) e^{x} " data-equation-content=" \displaystyle f'=(x^{2} - 2 x + 6)(\left(- 4 x - 2\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) e^{x})+(\left(- 2 x^{2} - 2 x + 1\right) e^{x})(2 x - 2)=\left(- 4 x - 2\right) \left(x^{2} - 2 x + 6\right) e^{x} + \left(2 x - 2\right) \left(- 2 x^{2} - 2 x + 1\right) e^{x} + \left(- 2 x^{2} - 2 x + 1\right) \left(x^{2} - 2 x + 6\right) e^{x} " /> </p> </p>