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

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

\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.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (\log{\left(x \right)})(\cos{\left(x \right)})(- 2 x - 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%202%20x%20-%208%29%20" alt="LaTeX:  \displaystyle y = (\log{\left(x \right)})(\cos{\left(x \right)})(- 2 x - 8) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(\cos{\left(x \right)})(- 2 x - 8) " /> .</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 - 8\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(- 2 x - 8\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- 2 x - 8\right) \cos{\left(x \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 - 8\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(- 2 x - 8\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- 2 x - 8\right) \cos{\left(x \right)} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " data-equation-content=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=- 2 x - 8 \implies g'=-2 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%202%20x%20-%208%20%5Cimplies%20g%27%3D-2%20" alt="LaTeX:  \displaystyle g=- 2 x - 8 \implies g'=-2 " data-equation-content=" \displaystyle g=- 2 x - 8 \implies g'=-2 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(- 2 x - 8)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(-2) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%202%20x%20-%208%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28-2%29%20" alt="LaTeX:  \displaystyle g'=(- 2 x - 8)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(-2) " data-equation-content=" \displaystyle g'=(- 2 x - 8)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(-2) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(- \left(- 2 x - 8\right) \sin{\left(x \right)} - 2 \cos{\left(x \right)})+(\left(- 2 x - 8\right) \cos{\left(x \right)})(\frac{1}{x})=- \left(- 2 x - 8\right) \log{\left(x \right)} \sin{\left(x \right)} - 2 \log{\left(x \right)} \cos{\left(x \right)} + \frac{\left(- 2 x - 8\right) \cos{\left(x \right)}}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%20%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20-%202%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D-%20%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20-%202%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%202%20x%20-%208%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(\log{\left(x \right)})(- \left(- 2 x - 8\right) \sin{\left(x \right)} - 2 \cos{\left(x \right)})+(\left(- 2 x - 8\right) \cos{\left(x \right)})(\frac{1}{x})=- \left(- 2 x - 8\right) \log{\left(x \right)} \sin{\left(x \right)} - 2 \log{\left(x \right)} \cos{\left(x \right)} + \frac{\left(- 2 x - 8\right) \cos{\left(x \right)}}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(- \left(- 2 x - 8\right) \sin{\left(x \right)} - 2 \cos{\left(x \right)})+(\left(- 2 x - 8\right) \cos{\left(x \right)})(\frac{1}{x})=- \left(- 2 x - 8\right) \log{\left(x \right)} \sin{\left(x \right)} - 2 \log{\left(x \right)} \cos{\left(x \right)} + \frac{\left(- 2 x - 8\right) \cos{\left(x \right)}}{x} " /> </p> </p>