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Identifying \(\displaystyle f=\log{\left(x \right)}\) and \(\displaystyle g=\left(- 7 x - 8\right) \left(- x - 3\right)\) and using the product rule with \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\). This leaves g as \(\displaystyle g = \left(- 7 x - 8\right) \left(- x - 3\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 7 x - 8 \implies f'=-7\) and \(\displaystyle g=- x - 3 \implies g'=-1\). Popping up a level gives \(\displaystyle g'=(- x - 3)(-7)+(- 7 x - 8)(-1)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\log{\left(x \right)})(14 x + 29)+(\left(- 7 x - 8\right) \left(- x - 3\right))(\frac{1}{x})=\left(7 x + 8\right) \log{\left(x \right)} + \left(7 x + 21\right) \log{\left(x \right)} + \frac{\left(- 7 x - 8\right) \left(- x - 3\right)}{x}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (\log{\left(x \right)})(- 7 x - 8)(- x - 3)$.
    \soln{9cm}{Identifying $f=\log{\left(x \right)}$ and $g=\left(- 7 x - 8\right) \left(- x - 3\right)$ and using the product rule with $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$. This leaves g as $g = \left(- 7 x - 8\right) \left(- x - 3\right)$ which also requires the product rule. Pushing down in the new product rule $f=- 7 x - 8 \implies f'=-7$ and $g=- x - 3 \implies g'=-1$. Popping up a level gives $g'=(- x - 3)(-7)+(- 7 x - 8)(-1)$Popping up again (Back to the original problem) gives $f'=(\log{\left(x \right)})(14 x + 29)+(\left(- 7 x - 8\right) \left(- x - 3\right))(\frac{1}{x})=\left(7 x + 8\right) \log{\left(x \right)} + \left(7 x + 21\right) \log{\left(x \right)} + \frac{\left(- 7 x - 8\right) \left(- x - 3\right)}{x}$}

\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)})(- 7 x - 8)(- x - 3) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%207%20x%20-%208%29%28-%20x%20-%203%29%20" alt="LaTeX:  \displaystyle y = (\log{\left(x \right)})(- 7 x - 8)(- x - 3) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(- 7 x - 8)(- x - 3) " /> .</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(- 7 x - 8\right) \left(- x - 3\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%207%20x%20-%208%5Cright%29%20%5Cleft%28-%20x%20-%203%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(- 7 x - 8\right) \left(- x - 3\right) " data-equation-content=" \displaystyle g=\left(- 7 x - 8\right) \left(- x - 3\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(- 7 x - 8\right) \left(- x - 3\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%207%20x%20-%208%5Cright%29%20%5Cleft%28-%20x%20-%203%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(- 7 x - 8\right) \left(- x - 3\right) " data-equation-content=" \displaystyle g = \left(- 7 x - 8\right) \left(- x - 3\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=- 7 x - 8 \implies f'=-7 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%207%20x%20-%208%20%5Cimplies%20f%27%3D-7%20" alt="LaTeX:  \displaystyle f=- 7 x - 8 \implies f'=-7 " data-equation-content=" \displaystyle f=- 7 x - 8 \implies f'=-7 " />  and  <img class="equation_image" title=" \displaystyle g=- x - 3 \implies g'=-1 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%20x%20-%203%20%5Cimplies%20g%27%3D-1%20" alt="LaTeX:  \displaystyle g=- x - 3 \implies g'=-1 " data-equation-content=" \displaystyle g=- x - 3 \implies g'=-1 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(- x - 3)(-7)+(- 7 x - 8)(-1) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%20x%20-%203%29%28-7%29%2B%28-%207%20x%20-%208%29%28-1%29%20" alt="LaTeX:  \displaystyle g'=(- x - 3)(-7)+(- 7 x - 8)(-1) " data-equation-content=" \displaystyle g'=(- x - 3)(-7)+(- 7 x - 8)(-1) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(14 x + 29)+(\left(- 7 x - 8\right) \left(- x - 3\right))(\frac{1}{x})=\left(7 x + 8\right) \log{\left(x \right)} + \left(7 x + 21\right) \log{\left(x \right)} + \frac{\left(- 7 x - 8\right) \left(- x - 3\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2814%20x%20%2B%2029%29%2B%28%5Cleft%28-%207%20x%20-%208%5Cright%29%20%5Cleft%28-%20x%20-%203%5Cright%29%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D%5Cleft%287%20x%20%2B%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%287%20x%20%2B%2021%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%207%20x%20-%208%5Cright%29%20%5Cleft%28-%20x%20-%203%5Cright%29%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(\log{\left(x \right)})(14 x + 29)+(\left(- 7 x - 8\right) \left(- x - 3\right))(\frac{1}{x})=\left(7 x + 8\right) \log{\left(x \right)} + \left(7 x + 21\right) \log{\left(x \right)} + \frac{\left(- 7 x - 8\right) \left(- x - 3\right)}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(14 x + 29)+(\left(- 7 x - 8\right) \left(- x - 3\right))(\frac{1}{x})=\left(7 x + 8\right) \log{\left(x \right)} + \left(7 x + 21\right) \log{\left(x \right)} + \frac{\left(- 7 x - 8\right) \left(- x - 3\right)}{x} " /> </p> </p>