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Identifying \(\displaystyle f=8 x^{2} - 7 x + 2\) and \(\displaystyle g=\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=8 x^{2} - 7 x + 2 \implies f'=16 x - 7\). This leaves g as \(\displaystyle g = \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=7 x^{2} + 6 x + 3 \implies f'=14 x + 6\) and \(\displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x}\). Popping up a level gives \(\displaystyle g'=(\log{\left(x \right)})(14 x + 6)+(7 x^{2} + 6 x + 3)(\frac{1}{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(8 x^{2} - 7 x + 2)(\left(14 x + 6\right) \log{\left(x \right)} + \frac{7 x^{2} + 6 x + 3}{x})+(\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)})(16 x - 7)=\left(14 x + 6\right) \left(8 x^{2} - 7 x + 2\right) \log{\left(x \right)} + \left(16 x - 7\right) \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} + \frac{\left(7 x^{2} + 6 x + 3\right) \left(8 x^{2} - 7 x + 2\right)}{x}\)

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\begin{question}Find the derivative of $y = (8 x^{2} - 7 x + 2)(7 x^{2} + 6 x + 3)(\log{\left(x \right)})$.
    \soln{9cm}{Identifying $f=8 x^{2} - 7 x + 2$ and $g=\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)}$ and using the product rule with $f=8 x^{2} - 7 x + 2 \implies f'=16 x - 7$. This leaves g as $g = \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=7 x^{2} + 6 x + 3 \implies f'=14 x + 6$ and $g=\log{\left(x \right)} \implies g'=\frac{1}{x}$. Popping up a level gives $g'=(\log{\left(x \right)})(14 x + 6)+(7 x^{2} + 6 x + 3)(\frac{1}{x})$Popping up again (Back to the original problem) gives $f'=(8 x^{2} - 7 x + 2)(\left(14 x + 6\right) \log{\left(x \right)} + \frac{7 x^{2} + 6 x + 3}{x})+(\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)})(16 x - 7)=\left(14 x + 6\right) \left(8 x^{2} - 7 x + 2\right) \log{\left(x \right)} + \left(16 x - 7\right) \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} + \frac{\left(7 x^{2} + 6 x + 3\right) \left(8 x^{2} - 7 x + 2\right)}{x}$}

\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 = (8 x^{2} - 7 x + 2)(7 x^{2} + 6 x + 3)(\log{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%288%20x%5E%7B2%7D%20-%207%20x%20%2B%202%29%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX:  \displaystyle y = (8 x^{2} - 7 x + 2)(7 x^{2} + 6 x + 3)(\log{\left(x \right)}) " data-equation-content=" \displaystyle y = (8 x^{2} - 7 x + 2)(7 x^{2} + 6 x + 3)(\log{\left(x \right)}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=8 x^{2} - 7 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%5E%7B2%7D%20-%207%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f=8 x^{2} - 7 x + 2 " data-equation-content=" \displaystyle f=8 x^{2} - 7 x + 2 " />  and  <img class="equation_image" title=" \displaystyle g=\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=8 x^{2} - 7 x + 2 \implies f'=16 x - 7 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%5E%7B2%7D%20-%207%20x%20%2B%202%20%5Cimplies%20f%27%3D16%20x%20-%207%20" alt="LaTeX:  \displaystyle f=8 x^{2} - 7 x + 2 \implies f'=16 x - 7 " data-equation-content=" \displaystyle f=8 x^{2} - 7 x + 2 \implies f'=16 x - 7 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(7 x^{2} + 6 x + 3\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=7 x^{2} + 6 x + 3 \implies f'=14 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D7%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%20%5Cimplies%20f%27%3D14%20x%20%2B%206%20" alt="LaTeX:  \displaystyle f=7 x^{2} + 6 x + 3 \implies f'=14 x + 6 " data-equation-content=" \displaystyle f=7 x^{2} + 6 x + 3 \implies f'=14 x + 6 " />  and  <img class="equation_image" title=" \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20g%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " data-equation-content=" \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(\log{\left(x \right)})(14 x + 6)+(7 x^{2} + 6 x + 3)(\frac{1}{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2814%20x%20%2B%206%29%2B%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%29%28%5Cfrac%7B1%7D%7Bx%7D%29%20" alt="LaTeX:  \displaystyle g'=(\log{\left(x \right)})(14 x + 6)+(7 x^{2} + 6 x + 3)(\frac{1}{x}) " data-equation-content=" \displaystyle g'=(\log{\left(x \right)})(14 x + 6)+(7 x^{2} + 6 x + 3)(\frac{1}{x}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(8 x^{2} - 7 x + 2)(\left(14 x + 6\right) \log{\left(x \right)} + \frac{7 x^{2} + 6 x + 3}{x})+(\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)})(16 x - 7)=\left(14 x + 6\right) \left(8 x^{2} - 7 x + 2\right) \log{\left(x \right)} + \left(16 x - 7\right) \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} + \frac{\left(7 x^{2} + 6 x + 3\right) \left(8 x^{2} - 7 x + 2\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%288%20x%5E%7B2%7D%20-%207%20x%20%2B%202%29%28%5Cleft%2814%20x%20%2B%206%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B7%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%7D%7Bx%7D%29%2B%28%5Cleft%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2816%20x%20-%207%29%3D%5Cleft%2814%20x%20%2B%206%5Cright%29%20%5Cleft%288%20x%5E%7B2%7D%20-%207%20x%20%2B%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2816%20x%20-%207%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%287%20x%5E%7B2%7D%20%2B%206%20x%20%2B%203%5Cright%29%20%5Cleft%288%20x%5E%7B2%7D%20-%207%20x%20%2B%202%5Cright%29%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(8 x^{2} - 7 x + 2)(\left(14 x + 6\right) \log{\left(x \right)} + \frac{7 x^{2} + 6 x + 3}{x})+(\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)})(16 x - 7)=\left(14 x + 6\right) \left(8 x^{2} - 7 x + 2\right) \log{\left(x \right)} + \left(16 x - 7\right) \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} + \frac{\left(7 x^{2} + 6 x + 3\right) \left(8 x^{2} - 7 x + 2\right)}{x} " data-equation-content=" \displaystyle f'=(8 x^{2} - 7 x + 2)(\left(14 x + 6\right) \log{\left(x \right)} + \frac{7 x^{2} + 6 x + 3}{x})+(\left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)})(16 x - 7)=\left(14 x + 6\right) \left(8 x^{2} - 7 x + 2\right) \log{\left(x \right)} + \left(16 x - 7\right) \left(7 x^{2} + 6 x + 3\right) \log{\left(x \right)} + \frac{\left(7 x^{2} + 6 x + 3\right) \left(8 x^{2} - 7 x + 2\right)}{x} " /> </p> </p>