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Identifying \(\displaystyle f=\log{\left(x \right)}\) and \(\displaystyle g=\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)\) and using the product rule with \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\). This leaves g as \(\displaystyle g = \left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=7 x^{2} + 8 x - 6 \implies f'=14 x + 8\) and \(\displaystyle g=2 x^{2} + 7 x - 2 \implies g'=4 x + 7\). Popping up a level gives \(\displaystyle g'=(2 x^{2} + 7 x - 2)(14 x + 8)+(7 x^{2} + 8 x - 6)(4 x + 7)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\log{\left(x \right)})(\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right))+(\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right))(\frac{1}{x})=\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) \log{\left(x \right)} + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right) \log{\left(x \right)} + \frac{\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)}{x}\)

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\begin{question}Find the derivative of $y = (\log{\left(x \right)})(7 x^{2} + 8 x - 6)(2 x^{2} + 7 x - 2)$.
    \soln{9cm}{Identifying $f=\log{\left(x \right)}$ and $g=\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)$ and using the product rule with $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$. This leaves g as $g = \left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)$ which also requires the product rule. Pushing down in the new product rule $f=7 x^{2} + 8 x - 6 \implies f'=14 x + 8$ and $g=2 x^{2} + 7 x - 2 \implies g'=4 x + 7$. Popping up a level gives $g'=(2 x^{2} + 7 x - 2)(14 x + 8)+(7 x^{2} + 8 x - 6)(4 x + 7)$Popping up again (Back to the original problem) gives $f'=(\log{\left(x \right)})(\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right))+(\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right))(\frac{1}{x})=\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) \log{\left(x \right)} + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right) \log{\left(x \right)} + \frac{\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)}{x}$}

\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 = (\log{\left(x \right)})(7 x^{2} + 8 x - 6)(2 x^{2} + 7 x - 2) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%29%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%29%20" alt="LaTeX:  \displaystyle y = (\log{\left(x \right)})(7 x^{2} + 8 x - 6)(2 x^{2} + 7 x - 2) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(7 x^{2} + 8 x - 6)(2 x^{2} + 7 x - 2) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} " data-equation-content=" \displaystyle f=\log{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " data-equation-content=" \displaystyle g=\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " />  and using the product rule with  <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} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " data-equation-content=" \displaystyle g = \left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=7 x^{2} + 8 x - 6 \implies f'=14 x + 8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D7%20x%5E%7B2%7D%20%2B%208%20x%20-%206%20%5Cimplies%20f%27%3D14%20x%20%2B%208%20" alt="LaTeX:  \displaystyle f=7 x^{2} + 8 x - 6 \implies f'=14 x + 8 " data-equation-content=" \displaystyle f=7 x^{2} + 8 x - 6 \implies f'=14 x + 8 " />  and  <img class="equation_image" title=" \displaystyle g=2 x^{2} + 7 x - 2 \implies g'=4 x + 7 " src="/equation_images/%20%5Cdisplaystyle%20g%3D2%20x%5E%7B2%7D%20%2B%207%20x%20-%202%20%5Cimplies%20g%27%3D4%20x%20%2B%207%20" alt="LaTeX:  \displaystyle g=2 x^{2} + 7 x - 2 \implies g'=4 x + 7 " data-equation-content=" \displaystyle g=2 x^{2} + 7 x - 2 \implies g'=4 x + 7 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(2 x^{2} + 7 x - 2)(14 x + 8)+(7 x^{2} + 8 x - 6)(4 x + 7) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%29%2814%20x%20%2B%208%29%2B%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%29%284%20x%20%2B%207%29%20" alt="LaTeX:  \displaystyle g'=(2 x^{2} + 7 x - 2)(14 x + 8)+(7 x^{2} + 8 x - 6)(4 x + 7) " data-equation-content=" \displaystyle g'=(2 x^{2} + 7 x - 2)(14 x + 8)+(7 x^{2} + 8 x - 6)(4 x + 7) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right))+(\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right))(\frac{1}{x})=\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) \log{\left(x \right)} + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right) \log{\left(x \right)} + \frac{\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%284%20x%20%2B%207%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%20%2B%20%5Cleft%2814%20x%20%2B%208%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%29%2B%28%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D%5Cleft%284%20x%20%2B%207%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2814%20x%20%2B%208%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%282%20x%5E%7B2%7D%20%2B%207%20x%20-%202%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%208%20x%20-%206%5Cright%29%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(\log{\left(x \right)})(\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right))+(\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right))(\frac{1}{x})=\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) \log{\left(x \right)} + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right) \log{\left(x \right)} + \frac{\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right))+(\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right))(\frac{1}{x})=\left(4 x + 7\right) \left(7 x^{2} + 8 x - 6\right) \log{\left(x \right)} + \left(14 x + 8\right) \left(2 x^{2} + 7 x - 2\right) \log{\left(x \right)} + \frac{\left(2 x^{2} + 7 x - 2\right) \left(7 x^{2} + 8 x - 6\right)}{x} " /> </p> </p>