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Find the derivative of \(\displaystyle y = (- 7 x^{3} - x^{2} - x - 2)(\log{\left(x \right)})(5 x^{3} + 6 x^{2} + x + 7)\).
Identifying \(\displaystyle f=- 7 x^{3} - x^{2} - x - 2\) and \(\displaystyle g=\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=- 7 x^{3} - x^{2} - x - 2 \implies f'=- 21 x^{2} - 2 x - 1\). This leaves g as \(\displaystyle g = \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)}\) 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 \(\displaystyle g=5 x^{3} + 6 x^{2} + x + 7 \implies g'=15 x^{2} + 12 x + 1\). Popping up a level gives \(\displaystyle g'=(5 x^{3} + 6 x^{2} + x + 7)(\frac{1}{x})+(\log{\left(x \right)})(15 x^{2} + 12 x + 1)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- 7 x^{3} - x^{2} - x - 2)(\left(15 x^{2} + 12 x + 1\right) \log{\left(x \right)} + \frac{5 x^{3} + 6 x^{2} + x + 7}{x})+(\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)})(- 21 x^{2} - 2 x - 1)=\left(- 21 x^{2} - 2 x - 1\right) \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} + \left(15 x^{2} + 12 x + 1\right) \left(- 7 x^{3} - x^{2} - x - 2\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - x^{2} - x - 2\right) \left(5 x^{3} + 6 x^{2} + x + 7\right)}{x}\)
\begin{question}Find the derivative of $y = (- 7 x^{3} - x^{2} - x - 2)(\log{\left(x \right)})(5 x^{3} + 6 x^{2} + x + 7)$.
\soln{9cm}{Identifying $f=- 7 x^{3} - x^{2} - x - 2$ and $g=\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)}$ and using the product rule with $f=- 7 x^{3} - x^{2} - x - 2 \implies f'=- 21 x^{2} - 2 x - 1$. This leaves g as $g = \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$ and $g=5 x^{3} + 6 x^{2} + x + 7 \implies g'=15 x^{2} + 12 x + 1$. Popping up a level gives $g'=(5 x^{3} + 6 x^{2} + x + 7)(\frac{1}{x})+(\log{\left(x \right)})(15 x^{2} + 12 x + 1)$Popping up again (Back to the original problem) gives $f'=(- 7 x^{3} - x^{2} - x - 2)(\left(15 x^{2} + 12 x + 1\right) \log{\left(x \right)} + \frac{5 x^{3} + 6 x^{2} + x + 7}{x})+(\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)})(- 21 x^{2} - 2 x - 1)=\left(- 21 x^{2} - 2 x - 1\right) \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} + \left(15 x^{2} + 12 x + 1\right) \left(- 7 x^{3} - x^{2} - x - 2\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - x^{2} - x - 2\right) \left(5 x^{3} + 6 x^{2} + x + 7\right)}{x}$}
\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}<p> <p>Find the derivative of <img class="equation_image" title=" \displaystyle y = (- 7 x^{3} - x^{2} - x - 2)(\log{\left(x \right)})(5 x^{3} + 6 x^{2} + x + 7) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%29%20" alt="LaTeX: \displaystyle y = (- 7 x^{3} - x^{2} - x - 2)(\log{\left(x \right)})(5 x^{3} + 6 x^{2} + x + 7) " data-equation-content=" \displaystyle y = (- 7 x^{3} - x^{2} - x - 2)(\log{\left(x \right)})(5 x^{3} + 6 x^{2} + x + 7) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=- 7 x^{3} - x^{2} - x - 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%20" alt="LaTeX: \displaystyle f=- 7 x^{3} - x^{2} - x - 2 " data-equation-content=" \displaystyle f=- 7 x^{3} - x^{2} - x - 2 " /> and <img class="equation_image" title=" \displaystyle g=\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=- 7 x^{3} - x^{2} - x - 2 \implies f'=- 21 x^{2} - 2 x - 1 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%20%5Cimplies%20f%27%3D-%2021%20x%5E%7B2%7D%20-%202%20x%20-%201%20" alt="LaTeX: \displaystyle f=- 7 x^{3} - x^{2} - x - 2 \implies f'=- 21 x^{2} - 2 x - 1 " data-equation-content=" \displaystyle f=- 7 x^{3} - x^{2} - x - 2 \implies f'=- 21 x^{2} - 2 x - 1 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(5 x^{3} + 6 x^{2} + x + 7\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=5 x^{3} + 6 x^{2} + x + 7 \implies g'=15 x^{2} + 12 x + 1 " src="/equation_images/%20%5Cdisplaystyle%20g%3D5%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%20%5Cimplies%20g%27%3D15%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%201%20" alt="LaTeX: \displaystyle g=5 x^{3} + 6 x^{2} + x + 7 \implies g'=15 x^{2} + 12 x + 1 " data-equation-content=" \displaystyle g=5 x^{3} + 6 x^{2} + x + 7 \implies g'=15 x^{2} + 12 x + 1 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(5 x^{3} + 6 x^{2} + x + 7)(\frac{1}{x})+(\log{\left(x \right)})(15 x^{2} + 12 x + 1) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%29%28%5Cfrac%7B1%7D%7Bx%7D%29%2B%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2815%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%201%29%20" alt="LaTeX: \displaystyle g'=(5 x^{3} + 6 x^{2} + x + 7)(\frac{1}{x})+(\log{\left(x \right)})(15 x^{2} + 12 x + 1) " data-equation-content=" \displaystyle g'=(5 x^{3} + 6 x^{2} + x + 7)(\frac{1}{x})+(\log{\left(x \right)})(15 x^{2} + 12 x + 1) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(- 7 x^{3} - x^{2} - x - 2)(\left(15 x^{2} + 12 x + 1\right) \log{\left(x \right)} + \frac{5 x^{3} + 6 x^{2} + x + 7}{x})+(\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)})(- 21 x^{2} - 2 x - 1)=\left(- 21 x^{2} - 2 x - 1\right) \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} + \left(15 x^{2} + 12 x + 1\right) \left(- 7 x^{3} - x^{2} - x - 2\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - x^{2} - x - 2\right) \left(5 x^{3} + 6 x^{2} + x + 7\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%29%28%5Cleft%2815%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B5%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%7D%7Bx%7D%29%2B%28%5Cleft%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%2021%20x%5E%7B2%7D%20-%202%20x%20-%201%29%3D%5Cleft%28-%2021%20x%5E%7B2%7D%20-%202%20x%20-%201%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2815%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%201%5Cright%29%20%5Cleft%28-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%207%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%20x%20-%202%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%20x%20%2B%207%5Cright%29%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(- 7 x^{3} - x^{2} - x - 2)(\left(15 x^{2} + 12 x + 1\right) \log{\left(x \right)} + \frac{5 x^{3} + 6 x^{2} + x + 7}{x})+(\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)})(- 21 x^{2} - 2 x - 1)=\left(- 21 x^{2} - 2 x - 1\right) \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} + \left(15 x^{2} + 12 x + 1\right) \left(- 7 x^{3} - x^{2} - x - 2\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - x^{2} - x - 2\right) \left(5 x^{3} + 6 x^{2} + x + 7\right)}{x} " data-equation-content=" \displaystyle f'=(- 7 x^{3} - x^{2} - x - 2)(\left(15 x^{2} + 12 x + 1\right) \log{\left(x \right)} + \frac{5 x^{3} + 6 x^{2} + x + 7}{x})+(\left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)})(- 21 x^{2} - 2 x - 1)=\left(- 21 x^{2} - 2 x - 1\right) \left(5 x^{3} + 6 x^{2} + x + 7\right) \log{\left(x \right)} + \left(15 x^{2} + 12 x + 1\right) \left(- 7 x^{3} - x^{2} - x - 2\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - x^{2} - x - 2\right) \left(5 x^{3} + 6 x^{2} + x + 7\right)}{x} " /> </p> </p>