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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (6 x^{3} - 6 x^{2} - 4 x - 4)(- 8 x^{3} - 6 x^{2} - x + 9)(\log{\left(x \right)})\).
Identifying \(\displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4\) and \(\displaystyle g=\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 \implies f'=18 x^{2} - 12 x - 4\). This leaves g as \(\displaystyle g = \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 8 x^{3} - 6 x^{2} - x + 9 \implies f'=- 24 x^{2} - 12 x - 1\) and \(\displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x}\). Popping up a level gives \(\displaystyle g'=(\log{\left(x \right)})(- 24 x^{2} - 12 x - 1)+(- 8 x^{3} - 6 x^{2} - x + 9)(\frac{1}{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(6 x^{3} - 6 x^{2} - 4 x - 4)(\left(- 24 x^{2} - 12 x - 1\right) \log{\left(x \right)} + \frac{- 8 x^{3} - 6 x^{2} - x + 9}{x})+(\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)})(18 x^{2} - 12 x - 4)=\left(- 24 x^{2} - 12 x - 1\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right) \log{\left(x \right)} + \left(18 x^{2} - 12 x - 4\right) \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} + \frac{\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right)}{x}\)
\begin{question}Find the derivative of $y = (6 x^{3} - 6 x^{2} - 4 x - 4)(- 8 x^{3} - 6 x^{2} - x + 9)(\log{\left(x \right)})$.
\soln{9cm}{Identifying $f=6 x^{3} - 6 x^{2} - 4 x - 4$ and $g=\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)}$ and using the product rule with $f=6 x^{3} - 6 x^{2} - 4 x - 4 \implies f'=18 x^{2} - 12 x - 4$. This leaves g as $g = \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=- 8 x^{3} - 6 x^{2} - x + 9 \implies f'=- 24 x^{2} - 12 x - 1$ and $g=\log{\left(x \right)} \implies g'=\frac{1}{x}$. Popping up a level gives $g'=(\log{\left(x \right)})(- 24 x^{2} - 12 x - 1)+(- 8 x^{3} - 6 x^{2} - x + 9)(\frac{1}{x})$Popping up again (Back to the original problem) gives $f'=(6 x^{3} - 6 x^{2} - 4 x - 4)(\left(- 24 x^{2} - 12 x - 1\right) \log{\left(x \right)} + \frac{- 8 x^{3} - 6 x^{2} - x + 9}{x})+(\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)})(18 x^{2} - 12 x - 4)=\left(- 24 x^{2} - 12 x - 1\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right) \log{\left(x \right)} + \left(18 x^{2} - 12 x - 4\right) \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} + \frac{\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right)}{x}$}
\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 = (6 x^{3} - 6 x^{2} - 4 x - 4)(- 8 x^{3} - 6 x^{2} - x + 9)(\log{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%286%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%29%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX: \displaystyle y = (6 x^{3} - 6 x^{2} - 4 x - 4)(- 8 x^{3} - 6 x^{2} - x + 9)(\log{\left(x \right)}) " data-equation-content=" \displaystyle y = (6 x^{3} - 6 x^{2} - 4 x - 4)(- 8 x^{3} - 6 x^{2} - x + 9)(\log{\left(x \right)}) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%3D6%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%20" alt="LaTeX: \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 " data-equation-content=" \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 " /> and <img class="equation_image" title=" \displaystyle g=\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 \implies f'=18 x^{2} - 12 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%3D6%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%20%5Cimplies%20f%27%3D18%20x%5E%7B2%7D%20-%2012%20x%20-%204%20" alt="LaTeX: \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 \implies f'=18 x^{2} - 12 x - 4 " data-equation-content=" \displaystyle f=6 x^{3} - 6 x^{2} - 4 x - 4 \implies f'=18 x^{2} - 12 x - 4 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- 8 x^{3} - 6 x^{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=- 8 x^{3} - 6 x^{2} - x + 9 \implies f'=- 24 x^{2} - 12 x - 1 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%20%5Cimplies%20f%27%3D-%2024%20x%5E%7B2%7D%20-%2012%20x%20-%201%20" alt="LaTeX: \displaystyle f=- 8 x^{3} - 6 x^{2} - x + 9 \implies f'=- 24 x^{2} - 12 x - 1 " data-equation-content=" \displaystyle f=- 8 x^{3} - 6 x^{2} - x + 9 \implies f'=- 24 x^{2} - 12 x - 1 " /> and <img class="equation_image" title=" \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20g%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX: \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " data-equation-content=" \displaystyle g=\log{\left(x \right)} \implies g'=\frac{1}{x} " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(\log{\left(x \right)})(- 24 x^{2} - 12 x - 1)+(- 8 x^{3} - 6 x^{2} - x + 9)(\frac{1}{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%2024%20x%5E%7B2%7D%20-%2012%20x%20-%201%29%2B%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%29%28%5Cfrac%7B1%7D%7Bx%7D%29%20" alt="LaTeX: \displaystyle g'=(\log{\left(x \right)})(- 24 x^{2} - 12 x - 1)+(- 8 x^{3} - 6 x^{2} - x + 9)(\frac{1}{x}) " data-equation-content=" \displaystyle g'=(\log{\left(x \right)})(- 24 x^{2} - 12 x - 1)+(- 8 x^{3} - 6 x^{2} - x + 9)(\frac{1}{x}) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(6 x^{3} - 6 x^{2} - 4 x - 4)(\left(- 24 x^{2} - 12 x - 1\right) \log{\left(x \right)} + \frac{- 8 x^{3} - 6 x^{2} - x + 9}{x})+(\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)})(18 x^{2} - 12 x - 4)=\left(- 24 x^{2} - 12 x - 1\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right) \log{\left(x \right)} + \left(18 x^{2} - 12 x - 4\right) \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} + \frac{\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%286%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%29%28%5Cleft%28-%2024%20x%5E%7B2%7D%20-%2012%20x%20-%201%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%7D%7Bx%7D%29%2B%28%5Cleft%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2818%20x%5E%7B2%7D%20-%2012%20x%20-%204%29%3D%5Cleft%28-%2024%20x%5E%7B2%7D%20-%2012%20x%20-%201%5Cright%29%20%5Cleft%286%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2818%20x%5E%7B2%7D%20-%2012%20x%20-%204%5Cright%29%20%5Cleft%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%208%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%209%5Cright%29%20%5Cleft%286%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%204%20x%20-%204%5Cright%29%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(6 x^{3} - 6 x^{2} - 4 x - 4)(\left(- 24 x^{2} - 12 x - 1\right) \log{\left(x \right)} + \frac{- 8 x^{3} - 6 x^{2} - x + 9}{x})+(\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)})(18 x^{2} - 12 x - 4)=\left(- 24 x^{2} - 12 x - 1\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right) \log{\left(x \right)} + \left(18 x^{2} - 12 x - 4\right) \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} + \frac{\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right)}{x} " data-equation-content=" \displaystyle f'=(6 x^{3} - 6 x^{2} - 4 x - 4)(\left(- 24 x^{2} - 12 x - 1\right) \log{\left(x \right)} + \frac{- 8 x^{3} - 6 x^{2} - x + 9}{x})+(\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)})(18 x^{2} - 12 x - 4)=\left(- 24 x^{2} - 12 x - 1\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right) \log{\left(x \right)} + \left(18 x^{2} - 12 x - 4\right) \left(- 8 x^{3} - 6 x^{2} - x + 9\right) \log{\left(x \right)} + \frac{\left(- 8 x^{3} - 6 x^{2} - x + 9\right) \left(6 x^{3} - 6 x^{2} - 4 x - 4\right)}{x} " /> </p> </p>