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Identifying \(\displaystyle f=9 x^{2} - 9 x - 4\) and \(\displaystyle g=\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)}\) and using the product rule with \(\displaystyle f=9 x^{2} - 9 x - 4 \implies f'=18 x - 9\). This leaves g as \(\displaystyle g = \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\) and \(\displaystyle g=- 2 x^{2} - 9 x - 2 \implies g'=- 4 x - 9\). Popping up a level gives \(\displaystyle g'=(- 2 x^{2} - 9 x - 2)(\cos{\left(x \right)})+(\sin{\left(x \right)})(- 4 x - 9)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(9 x^{2} - 9 x - 4)(\left(- 4 x - 9\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \cos{\left(x \right)})+(\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)})(18 x - 9)=\left(- 4 x - 9\right) \left(9 x^{2} - 9 x - 4\right) \sin{\left(x \right)} + \left(18 x - 9\right) \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \left(9 x^{2} - 9 x - 4\right) \cos{\left(x \right)}\)

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\begin{question}Find the derivative of $y = (9 x^{2} - 9 x - 4)(\sin{\left(x \right)})(- 2 x^{2} - 9 x - 2)$.
    \soln{9cm}{Identifying $f=9 x^{2} - 9 x - 4$ and $g=\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)}$ and using the product rule with $f=9 x^{2} - 9 x - 4 \implies f'=18 x - 9$. This leaves g as $g = \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$ and $g=- 2 x^{2} - 9 x - 2 \implies g'=- 4 x - 9$. Popping up a level gives $g'=(- 2 x^{2} - 9 x - 2)(\cos{\left(x \right)})+(\sin{\left(x \right)})(- 4 x - 9)$Popping up again (Back to the original problem) gives $f'=(9 x^{2} - 9 x - 4)(\left(- 4 x - 9\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \cos{\left(x \right)})+(\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)})(18 x - 9)=\left(- 4 x - 9\right) \left(9 x^{2} - 9 x - 4\right) \sin{\left(x \right)} + \left(18 x - 9\right) \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \left(9 x^{2} - 9 x - 4\right) \cos{\left(x \right)}$}

\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 = (9 x^{2} - 9 x - 4)(\sin{\left(x \right)})(- 2 x^{2} - 9 x - 2) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%289%20x%5E%7B2%7D%20-%209%20x%20-%204%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%29%20" alt="LaTeX:  \displaystyle y = (9 x^{2} - 9 x - 4)(\sin{\left(x \right)})(- 2 x^{2} - 9 x - 2) " data-equation-content=" \displaystyle y = (9 x^{2} - 9 x - 4)(\sin{\left(x \right)})(- 2 x^{2} - 9 x - 2) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=9 x^{2} - 9 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%3D9%20x%5E%7B2%7D%20-%209%20x%20-%204%20" alt="LaTeX:  \displaystyle f=9 x^{2} - 9 x - 4 " data-equation-content=" \displaystyle f=9 x^{2} - 9 x - 4 " />  and  <img class="equation_image" title=" \displaystyle g=\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=9 x^{2} - 9 x - 4 \implies f'=18 x - 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D9%20x%5E%7B2%7D%20-%209%20x%20-%204%20%5Cimplies%20f%27%3D18%20x%20-%209%20" alt="LaTeX:  \displaystyle f=9 x^{2} - 9 x - 4 \implies f'=18 x - 9 " data-equation-content=" \displaystyle f=9 x^{2} - 9 x - 4 \implies f'=18 x - 9 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " data-equation-content=" \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=- 2 x^{2} - 9 x - 2 \implies g'=- 4 x - 9 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%20%5Cimplies%20g%27%3D-%204%20x%20-%209%20" alt="LaTeX:  \displaystyle g=- 2 x^{2} - 9 x - 2 \implies g'=- 4 x - 9 " data-equation-content=" \displaystyle g=- 2 x^{2} - 9 x - 2 \implies g'=- 4 x - 9 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(- 2 x^{2} - 9 x - 2)(\cos{\left(x \right)})+(\sin{\left(x \right)})(- 4 x - 9) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%204%20x%20-%209%29%20" alt="LaTeX:  \displaystyle g'=(- 2 x^{2} - 9 x - 2)(\cos{\left(x \right)})+(\sin{\left(x \right)})(- 4 x - 9) " data-equation-content=" \displaystyle g'=(- 2 x^{2} - 9 x - 2)(\cos{\left(x \right)})+(\sin{\left(x \right)})(- 4 x - 9) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(9 x^{2} - 9 x - 4)(\left(- 4 x - 9\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \cos{\left(x \right)})+(\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)})(18 x - 9)=\left(- 4 x - 9\right) \left(9 x^{2} - 9 x - 4\right) \sin{\left(x \right)} + \left(18 x - 9\right) \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \left(9 x^{2} - 9 x - 4\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%289%20x%5E%7B2%7D%20-%209%20x%20-%204%29%28%5Cleft%28-%204%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2818%20x%20-%209%29%3D%5Cleft%28-%204%20x%20-%209%5Cright%29%20%5Cleft%289%20x%5E%7B2%7D%20-%209%20x%20-%204%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2818%20x%20-%209%5Cright%29%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%202%20x%5E%7B2%7D%20-%209%20x%20-%202%5Cright%29%20%5Cleft%289%20x%5E%7B2%7D%20-%209%20x%20-%204%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(9 x^{2} - 9 x - 4)(\left(- 4 x - 9\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \cos{\left(x \right)})+(\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)})(18 x - 9)=\left(- 4 x - 9\right) \left(9 x^{2} - 9 x - 4\right) \sin{\left(x \right)} + \left(18 x - 9\right) \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \left(9 x^{2} - 9 x - 4\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(9 x^{2} - 9 x - 4)(\left(- 4 x - 9\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \cos{\left(x \right)})+(\left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)})(18 x - 9)=\left(- 4 x - 9\right) \left(9 x^{2} - 9 x - 4\right) \sin{\left(x \right)} + \left(18 x - 9\right) \left(- 2 x^{2} - 9 x - 2\right) \sin{\left(x \right)} + \left(- 2 x^{2} - 9 x - 2\right) \left(9 x^{2} - 9 x - 4\right) \cos{\left(x \right)} " /> </p> </p>