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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (- x^{3} - 8 x^{2} + 3 x - 3)(\log{\left(x \right)})(- 7 x^{3} - 2 x^{2} + 2 x + 8)\).
Identifying \(\displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3\) and \(\displaystyle g=\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 \implies f'=- 3 x^{2} - 16 x + 3\). This leaves g as \(\displaystyle g = \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)}\) 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 \(\displaystyle g=- 7 x^{3} - 2 x^{2} + 2 x + 8 \implies g'=- 21 x^{2} - 4 x + 2\). Popping up a level gives \(\displaystyle g'=(- 7 x^{3} - 2 x^{2} + 2 x + 8)(\frac{1}{x})+(\log{\left(x \right)})(- 21 x^{2} - 4 x + 2)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- x^{3} - 8 x^{2} + 3 x - 3)(\left(- 21 x^{2} - 4 x + 2\right) \log{\left(x \right)} + \frac{- 7 x^{3} - 2 x^{2} + 2 x + 8}{x})+(\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)})(- 3 x^{2} - 16 x + 3)=\left(- 21 x^{2} - 4 x + 2\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right) \log{\left(x \right)} + \left(- 3 x^{2} - 16 x + 3\right) \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right)}{x}\)
\begin{question}Find the derivative of $y = (- x^{3} - 8 x^{2} + 3 x - 3)(\log{\left(x \right)})(- 7 x^{3} - 2 x^{2} + 2 x + 8)$.
\soln{9cm}{Identifying $f=- x^{3} - 8 x^{2} + 3 x - 3$ and $g=\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)}$ and using the product rule with $f=- x^{3} - 8 x^{2} + 3 x - 3 \implies f'=- 3 x^{2} - 16 x + 3$. This leaves g as $g = \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$ and $g=- 7 x^{3} - 2 x^{2} + 2 x + 8 \implies g'=- 21 x^{2} - 4 x + 2$. Popping up a level gives $g'=(- 7 x^{3} - 2 x^{2} + 2 x + 8)(\frac{1}{x})+(\log{\left(x \right)})(- 21 x^{2} - 4 x + 2)$Popping up again (Back to the original problem) gives $f'=(- x^{3} - 8 x^{2} + 3 x - 3)(\left(- 21 x^{2} - 4 x + 2\right) \log{\left(x \right)} + \frac{- 7 x^{3} - 2 x^{2} + 2 x + 8}{x})+(\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)})(- 3 x^{2} - 16 x + 3)=\left(- 21 x^{2} - 4 x + 2\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right) \log{\left(x \right)} + \left(- 3 x^{2} - 16 x + 3\right) \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right)}{x}$}
\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 = (- x^{3} - 8 x^{2} + 3 x - 3)(\log{\left(x \right)})(- 7 x^{3} - 2 x^{2} + 2 x + 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%29%20" alt="LaTeX: \displaystyle y = (- x^{3} - 8 x^{2} + 3 x - 3)(\log{\left(x \right)})(- 7 x^{3} - 2 x^{2} + 2 x + 8) " data-equation-content=" \displaystyle y = (- x^{3} - 8 x^{2} + 3 x - 3)(\log{\left(x \right)})(- 7 x^{3} - 2 x^{2} + 2 x + 8) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%20" alt="LaTeX: \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 " data-equation-content=" \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 " /> and <img class="equation_image" title=" \displaystyle g=\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 \implies f'=- 3 x^{2} - 16 x + 3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%20%5Cimplies%20f%27%3D-%203%20x%5E%7B2%7D%20-%2016%20x%20%2B%203%20" alt="LaTeX: \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 \implies f'=- 3 x^{2} - 16 x + 3 " data-equation-content=" \displaystyle f=- x^{3} - 8 x^{2} + 3 x - 3 \implies f'=- 3 x^{2} - 16 x + 3 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\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=- 7 x^{3} - 2 x^{2} + 2 x + 8 \implies g'=- 21 x^{2} - 4 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%20%5Cimplies%20g%27%3D-%2021%20x%5E%7B2%7D%20-%204%20x%20%2B%202%20" alt="LaTeX: \displaystyle g=- 7 x^{3} - 2 x^{2} + 2 x + 8 \implies g'=- 21 x^{2} - 4 x + 2 " data-equation-content=" \displaystyle g=- 7 x^{3} - 2 x^{2} + 2 x + 8 \implies g'=- 21 x^{2} - 4 x + 2 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(- 7 x^{3} - 2 x^{2} + 2 x + 8)(\frac{1}{x})+(\log{\left(x \right)})(- 21 x^{2} - 4 x + 2) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%29%28%5Cfrac%7B1%7D%7Bx%7D%29%2B%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%2021%20x%5E%7B2%7D%20-%204%20x%20%2B%202%29%20" alt="LaTeX: \displaystyle g'=(- 7 x^{3} - 2 x^{2} + 2 x + 8)(\frac{1}{x})+(\log{\left(x \right)})(- 21 x^{2} - 4 x + 2) " data-equation-content=" \displaystyle g'=(- 7 x^{3} - 2 x^{2} + 2 x + 8)(\frac{1}{x})+(\log{\left(x \right)})(- 21 x^{2} - 4 x + 2) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(- x^{3} - 8 x^{2} + 3 x - 3)(\left(- 21 x^{2} - 4 x + 2\right) \log{\left(x \right)} + \frac{- 7 x^{3} - 2 x^{2} + 2 x + 8}{x})+(\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)})(- 3 x^{2} - 16 x + 3)=\left(- 21 x^{2} - 4 x + 2\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right) \log{\left(x \right)} + \left(- 3 x^{2} - 16 x + 3\right) \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%29%28%5Cleft%28-%2021%20x%5E%7B2%7D%20-%204%20x%20%2B%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%7D%7Bx%7D%29%2B%28%5Cleft%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%203%20x%5E%7B2%7D%20-%2016%20x%20%2B%203%29%3D%5Cleft%28-%2021%20x%5E%7B2%7D%20-%204%20x%20%2B%202%5Cright%29%20%5Cleft%28-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%203%20x%5E%7B2%7D%20-%2016%20x%20%2B%203%5Cright%29%20%5Cleft%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28-%207%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%202%20x%20%2B%208%5Cright%29%20%5Cleft%28-%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%203%5Cright%29%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(- x^{3} - 8 x^{2} + 3 x - 3)(\left(- 21 x^{2} - 4 x + 2\right) \log{\left(x \right)} + \frac{- 7 x^{3} - 2 x^{2} + 2 x + 8}{x})+(\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)})(- 3 x^{2} - 16 x + 3)=\left(- 21 x^{2} - 4 x + 2\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right) \log{\left(x \right)} + \left(- 3 x^{2} - 16 x + 3\right) \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right)}{x} " data-equation-content=" \displaystyle f'=(- x^{3} - 8 x^{2} + 3 x - 3)(\left(- 21 x^{2} - 4 x + 2\right) \log{\left(x \right)} + \frac{- 7 x^{3} - 2 x^{2} + 2 x + 8}{x})+(\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)})(- 3 x^{2} - 16 x + 3)=\left(- 21 x^{2} - 4 x + 2\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right) \log{\left(x \right)} + \left(- 3 x^{2} - 16 x + 3\right) \left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \log{\left(x \right)} + \frac{\left(- 7 x^{3} - 2 x^{2} + 2 x + 8\right) \left(- x^{3} - 8 x^{2} + 3 x - 3\right)}{x} " /> </p> </p>