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Find the derivative of \(\displaystyle y = (\log{\left(x \right)})(2 x^{3} - 8 x^{2} + 6 x + 2)(4 x^{3} + 5 x^{2} + 2 x - 8)\).
Identifying \(\displaystyle f=\log{\left(x \right)}\) and \(\displaystyle g=\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)\) and using the product rule with \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\). This leaves g as \(\displaystyle g = \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=2 x^{3} - 8 x^{2} + 6 x + 2 \implies f'=6 x^{2} - 16 x + 6\) and \(\displaystyle g=4 x^{3} + 5 x^{2} + 2 x - 8 \implies g'=12 x^{2} + 10 x + 2\). Popping up a level gives \(\displaystyle g'=(4 x^{3} + 5 x^{2} + 2 x - 8)(6 x^{2} - 16 x + 6)+(2 x^{3} - 8 x^{2} + 6 x + 2)(12 x^{2} + 10 x + 2)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\log{\left(x \right)})(\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right))+(\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right))(\frac{1}{x})=\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) \log{\left(x \right)} + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \log{\left(x \right)} + \frac{\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)}{x}\)
\begin{question}Find the derivative of $y = (\log{\left(x \right)})(2 x^{3} - 8 x^{2} + 6 x + 2)(4 x^{3} + 5 x^{2} + 2 x - 8)$.
\soln{9cm}{Identifying $f=\log{\left(x \right)}$ and $g=\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)$ and using the product rule with $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$. This leaves g as $g = \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)$ which also requires the product rule. Pushing down in the new product rule $f=2 x^{3} - 8 x^{2} + 6 x + 2 \implies f'=6 x^{2} - 16 x + 6$ and $g=4 x^{3} + 5 x^{2} + 2 x - 8 \implies g'=12 x^{2} + 10 x + 2$. Popping up a level gives $g'=(4 x^{3} + 5 x^{2} + 2 x - 8)(6 x^{2} - 16 x + 6)+(2 x^{3} - 8 x^{2} + 6 x + 2)(12 x^{2} + 10 x + 2)$Popping up again (Back to the original problem) gives $f'=(\log{\left(x \right)})(\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right))+(\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right))(\frac{1}{x})=\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) \log{\left(x \right)} + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \log{\left(x \right)} + \frac{\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)}{x}$}
\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 = (\log{\left(x \right)})(2 x^{3} - 8 x^{2} + 6 x + 2)(4 x^{3} + 5 x^{2} + 2 x - 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%29%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%29%20" alt="LaTeX: \displaystyle y = (\log{\left(x \right)})(2 x^{3} - 8 x^{2} + 6 x + 2)(4 x^{3} + 5 x^{2} + 2 x - 8) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(2 x^{3} - 8 x^{2} + 6 x + 2)(4 x^{3} + 5 x^{2} + 2 x - 8) " /> .</p> </p><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^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) " data-equation-content=" \displaystyle g=\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\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^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) " data-equation-content=" \displaystyle g = \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=2 x^{3} - 8 x^{2} + 6 x + 2 \implies f'=6 x^{2} - 16 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D2%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%20%5Cimplies%20f%27%3D6%20x%5E%7B2%7D%20-%2016%20x%20%2B%206%20" alt="LaTeX: \displaystyle f=2 x^{3} - 8 x^{2} + 6 x + 2 \implies f'=6 x^{2} - 16 x + 6 " data-equation-content=" \displaystyle f=2 x^{3} - 8 x^{2} + 6 x + 2 \implies f'=6 x^{2} - 16 x + 6 " /> and <img class="equation_image" title=" \displaystyle g=4 x^{3} + 5 x^{2} + 2 x - 8 \implies g'=12 x^{2} + 10 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20g%3D4%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%20%5Cimplies%20g%27%3D12%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%202%20" alt="LaTeX: \displaystyle g=4 x^{3} + 5 x^{2} + 2 x - 8 \implies g'=12 x^{2} + 10 x + 2 " data-equation-content=" \displaystyle g=4 x^{3} + 5 x^{2} + 2 x - 8 \implies g'=12 x^{2} + 10 x + 2 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(4 x^{3} + 5 x^{2} + 2 x - 8)(6 x^{2} - 16 x + 6)+(2 x^{3} - 8 x^{2} + 6 x + 2)(12 x^{2} + 10 x + 2) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%29%286%20x%5E%7B2%7D%20-%2016%20x%20%2B%206%29%2B%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%29%2812%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%202%29%20" alt="LaTeX: \displaystyle g'=(4 x^{3} + 5 x^{2} + 2 x - 8)(6 x^{2} - 16 x + 6)+(2 x^{3} - 8 x^{2} + 6 x + 2)(12 x^{2} + 10 x + 2) " data-equation-content=" \displaystyle g'=(4 x^{3} + 5 x^{2} + 2 x - 8)(6 x^{2} - 16 x + 6)+(2 x^{3} - 8 x^{2} + 6 x + 2)(12 x^{2} + 10 x + 2) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right))+(\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right))(\frac{1}{x})=\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) \log{\left(x \right)} + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \log{\left(x \right)} + \frac{\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%286%20x%5E%7B2%7D%20-%2016%20x%20%2B%206%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%20%2B%20%5Cleft%2812%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%202%5Cright%29%20%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%29%2B%28%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D%5Cleft%286%20x%5E%7B2%7D%20-%2016%20x%20%2B%206%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%202%5Cright%29%20%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%282%20x%5E%7B3%7D%20-%208%20x%5E%7B2%7D%20%2B%206%20x%20%2B%202%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%202%20x%20-%208%5Cright%29%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(\log{\left(x \right)})(\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right))+(\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right))(\frac{1}{x})=\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) \log{\left(x \right)} + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \log{\left(x \right)} + \frac{\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right))+(\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right))(\frac{1}{x})=\left(6 x^{2} - 16 x + 6\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right) \log{\left(x \right)} + \left(12 x^{2} + 10 x + 2\right) \left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \log{\left(x \right)} + \frac{\left(2 x^{3} - 8 x^{2} + 6 x + 2\right) \left(4 x^{3} + 5 x^{2} + 2 x - 8\right)}{x} " /> </p> </p>