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Identifying \(\displaystyle f=x^{2} + 8 x + 9\) and \(\displaystyle g=\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=x^{2} + 8 x + 9 \implies f'=2 x + 8\). This leaves g as \(\displaystyle g = \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\) and \(\displaystyle g=2 x^{2} + 6 x - 1 \implies g'=4 x + 6\). Popping up a level gives \(\displaystyle g'=(2 x^{2} + 6 x - 1)(\frac{1}{x})+(\log{\left(x \right)})(4 x + 6)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(x^{2} + 8 x + 9)(\left(4 x + 6\right) \log{\left(x \right)} + \frac{2 x^{2} + 6 x - 1}{x})+(\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)})(2 x + 8)=\left(2 x + 8\right) \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} + \left(4 x + 6\right) \left(x^{2} + 8 x + 9\right) \log{\left(x \right)} + \frac{\left(x^{2} + 8 x + 9\right) \left(2 x^{2} + 6 x - 1\right)}{x}\)

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\begin{question}Find the derivative of $y = (x^{2} + 8 x + 9)(\log{\left(x \right)})(2 x^{2} + 6 x - 1)$.
    \soln{9cm}{Identifying $f=x^{2} + 8 x + 9$ and $g=\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)}$ and using the product rule with $f=x^{2} + 8 x + 9 \implies f'=2 x + 8$. This leaves g as $g = \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$ and $g=2 x^{2} + 6 x - 1 \implies g'=4 x + 6$. Popping up a level gives $g'=(2 x^{2} + 6 x - 1)(\frac{1}{x})+(\log{\left(x \right)})(4 x + 6)$Popping up again (Back to the original problem) gives $f'=(x^{2} + 8 x + 9)(\left(4 x + 6\right) \log{\left(x \right)} + \frac{2 x^{2} + 6 x - 1}{x})+(\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)})(2 x + 8)=\left(2 x + 8\right) \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} + \left(4 x + 6\right) \left(x^{2} + 8 x + 9\right) \log{\left(x \right)} + \frac{\left(x^{2} + 8 x + 9\right) \left(2 x^{2} + 6 x - 1\right)}{x}$}

\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 = (x^{2} + 8 x + 9)(\log{\left(x \right)})(2 x^{2} + 6 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28x%5E%7B2%7D%20%2B%208%20x%20%2B%209%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%29%20" alt="LaTeX:  \displaystyle y = (x^{2} + 8 x + 9)(\log{\left(x \right)})(2 x^{2} + 6 x - 1) " data-equation-content=" \displaystyle y = (x^{2} + 8 x + 9)(\log{\left(x \right)})(2 x^{2} + 6 x - 1) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=x^{2} + 8 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B2%7D%20%2B%208%20x%20%2B%209%20" alt="LaTeX:  \displaystyle f=x^{2} + 8 x + 9 " data-equation-content=" \displaystyle f=x^{2} + 8 x + 9 " />  and  <img class="equation_image" title=" \displaystyle g=\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=x^{2} + 8 x + 9 \implies f'=2 x + 8 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B2%7D%20%2B%208%20x%20%2B%209%20%5Cimplies%20f%27%3D2%20x%20%2B%208%20" alt="LaTeX:  \displaystyle f=x^{2} + 8 x + 9 \implies f'=2 x + 8 " data-equation-content=" \displaystyle f=x^{2} + 8 x + 9 \implies f'=2 x + 8 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " data-equation-content=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " />  and  <img class="equation_image" title=" \displaystyle g=2 x^{2} + 6 x - 1 \implies g'=4 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20g%3D2%20x%5E%7B2%7D%20%2B%206%20x%20-%201%20%5Cimplies%20g%27%3D4%20x%20%2B%206%20" alt="LaTeX:  \displaystyle g=2 x^{2} + 6 x - 1 \implies g'=4 x + 6 " data-equation-content=" \displaystyle g=2 x^{2} + 6 x - 1 \implies g'=4 x + 6 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(2 x^{2} + 6 x - 1)(\frac{1}{x})+(\log{\left(x \right)})(4 x + 6) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%29%28%5Cfrac%7B1%7D%7Bx%7D%29%2B%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%284%20x%20%2B%206%29%20" alt="LaTeX:  \displaystyle g'=(2 x^{2} + 6 x - 1)(\frac{1}{x})+(\log{\left(x \right)})(4 x + 6) " data-equation-content=" \displaystyle g'=(2 x^{2} + 6 x - 1)(\frac{1}{x})+(\log{\left(x \right)})(4 x + 6) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(x^{2} + 8 x + 9)(\left(4 x + 6\right) \log{\left(x \right)} + \frac{2 x^{2} + 6 x - 1}{x})+(\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)})(2 x + 8)=\left(2 x + 8\right) \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} + \left(4 x + 6\right) \left(x^{2} + 8 x + 9\right) \log{\left(x \right)} + \frac{\left(x^{2} + 8 x + 9\right) \left(2 x^{2} + 6 x - 1\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28x%5E%7B2%7D%20%2B%208%20x%20%2B%209%29%28%5Cleft%284%20x%20%2B%206%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B2%20x%5E%7B2%7D%20%2B%206%20x%20-%201%7D%7Bx%7D%29%2B%28%5Cleft%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%282%20x%20%2B%208%29%3D%5Cleft%282%20x%20%2B%208%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%284%20x%20%2B%206%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20%2B%208%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28x%5E%7B2%7D%20%2B%208%20x%20%2B%209%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%206%20x%20-%201%5Cright%29%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(x^{2} + 8 x + 9)(\left(4 x + 6\right) \log{\left(x \right)} + \frac{2 x^{2} + 6 x - 1}{x})+(\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)})(2 x + 8)=\left(2 x + 8\right) \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} + \left(4 x + 6\right) \left(x^{2} + 8 x + 9\right) \log{\left(x \right)} + \frac{\left(x^{2} + 8 x + 9\right) \left(2 x^{2} + 6 x - 1\right)}{x} " data-equation-content=" \displaystyle f'=(x^{2} + 8 x + 9)(\left(4 x + 6\right) \log{\left(x \right)} + \frac{2 x^{2} + 6 x - 1}{x})+(\left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)})(2 x + 8)=\left(2 x + 8\right) \left(2 x^{2} + 6 x - 1\right) \log{\left(x \right)} + \left(4 x + 6\right) \left(x^{2} + 8 x + 9\right) \log{\left(x \right)} + \frac{\left(x^{2} + 8 x + 9\right) \left(2 x^{2} + 6 x - 1\right)}{x} " /> </p> </p>