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Identifying \(\displaystyle f=- 8 x - 1\) and \(\displaystyle g=\left(2 x - 9\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=- 8 x - 1 \implies f'=-8\). This leaves g as \(\displaystyle g = \left(2 x - 9\right) \log{\left(x \right)}\) 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 \(\displaystyle g=2 x - 9 \implies g'=2\). Popping up a level gives \(\displaystyle g'=(2 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(2)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- 8 x - 1)(2 \log{\left(x \right)} + \frac{2 x - 9}{x})+(\left(2 x - 9\right) \log{\left(x \right)})(-8)=\left(72 - 16 x\right) \log{\left(x \right)} + \left(- 16 x - 2\right) \log{\left(x \right)} + \frac{\left(- 8 x - 1\right) \left(2 x - 9\right)}{x}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (- 8 x - 1)(\log{\left(x \right)})(2 x - 9)$.
    \soln{9cm}{Identifying $f=- 8 x - 1$ and $g=\left(2 x - 9\right) \log{\left(x \right)}$ and using the product rule with $f=- 8 x - 1 \implies f'=-8$. This leaves g as $g = \left(2 x - 9\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$ and $g=2 x - 9 \implies g'=2$. Popping up a level gives $g'=(2 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(2)$Popping up again (Back to the original problem) gives $f'=(- 8 x - 1)(2 \log{\left(x \right)} + \frac{2 x - 9}{x})+(\left(2 x - 9\right) \log{\left(x \right)})(-8)=\left(72 - 16 x\right) \log{\left(x \right)} + \left(- 16 x - 2\right) \log{\left(x \right)} + \frac{\left(- 8 x - 1\right) \left(2 x - 9\right)}{x}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=- 8 x - 1 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%20-%201%20" alt="LaTeX:  \displaystyle f=- 8 x - 1 " data-equation-content=" \displaystyle f=- 8 x - 1 " />  and  <img class="equation_image" title=" \displaystyle g=\left(2 x - 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(2 x - 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(2 x - 9\right) \log{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=- 8 x - 1 \implies f'=-8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%20-%201%20%5Cimplies%20f%27%3D-8%20" alt="LaTeX:  \displaystyle f=- 8 x - 1 \implies f'=-8 " data-equation-content=" \displaystyle f=- 8 x - 1 \implies f'=-8 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(2 x - 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(2 x - 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(2 x - 9\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 - 9 \implies g'=2 " src="/equation_images/%20%5Cdisplaystyle%20g%3D2%20x%20-%209%20%5Cimplies%20g%27%3D2%20" alt="LaTeX:  \displaystyle g=2 x - 9 \implies g'=2 " data-equation-content=" \displaystyle g=2 x - 9 \implies g'=2 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(2 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(2) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%282%20x%20-%209%29%28%5Cfrac%7B1%7D%7Bx%7D%29%2B%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%282%29%20" alt="LaTeX:  \displaystyle g'=(2 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(2) " data-equation-content=" \displaystyle g'=(2 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(2) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(- 8 x - 1)(2 \log{\left(x \right)} + \frac{2 x - 9}{x})+(\left(2 x - 9\right) \log{\left(x \right)})(-8)=\left(72 - 16 x\right) \log{\left(x \right)} + \left(- 16 x - 2\right) \log{\left(x \right)} + \frac{\left(- 8 x - 1\right) \left(2 x - 9\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%208%20x%20-%201%29%282%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B2%20x%20-%209%7D%7Bx%7D%29%2B%28%5Cleft%282%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-8%29%3D%5Cleft%2872%20-%2016%20x%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%2016%20x%20-%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%208%20x%20-%201%5Cright%29%20%5Cleft%282%20x%20-%209%5Cright%29%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(- 8 x - 1)(2 \log{\left(x \right)} + \frac{2 x - 9}{x})+(\left(2 x - 9\right) \log{\left(x \right)})(-8)=\left(72 - 16 x\right) \log{\left(x \right)} + \left(- 16 x - 2\right) \log{\left(x \right)} + \frac{\left(- 8 x - 1\right) \left(2 x - 9\right)}{x} " data-equation-content=" \displaystyle f'=(- 8 x - 1)(2 \log{\left(x \right)} + \frac{2 x - 9}{x})+(\left(2 x - 9\right) \log{\left(x \right)})(-8)=\left(72 - 16 x\right) \log{\left(x \right)} + \left(- 16 x - 2\right) \log{\left(x \right)} + \frac{\left(- 8 x - 1\right) \left(2 x - 9\right)}{x} " /> </p> </p>