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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (4 - 6 x)(1 - x)(5 x + 8)\).
Identifying \(\displaystyle f=4 - 6 x\) and \(\displaystyle g=\left(1 - x\right) \left(5 x + 8\right)\) and using the product rule with \(\displaystyle f=4 - 6 x \implies f'=-6\). This leaves g as \(\displaystyle g = \left(1 - x\right) \left(5 x + 8\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=1 - x \implies f'=-1\) and \(\displaystyle g=5 x + 8 \implies g'=5\). Popping up a level gives \(\displaystyle g'=(5 x + 8)(-1)+(1 - x)(5)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(4 - 6 x)(- 10 x - 3)+(\left(1 - x\right) \left(5 x + 8\right))(-6)=5 \left(1 - x\right) \left(4 - 6 x\right) + \left(5 x + 8\right) \left(6 x - 6\right) + \left(5 x + 8\right) \left(6 x - 4\right)\)
\begin{question}Find the derivative of $y = (4 - 6 x)(1 - x)(5 x + 8)$.
\soln{9cm}{Identifying $f=4 - 6 x$ and $g=\left(1 - x\right) \left(5 x + 8\right)$ and using the product rule with $f=4 - 6 x \implies f'=-6$. This leaves g as $g = \left(1 - x\right) \left(5 x + 8\right)$ which also requires the product rule. Pushing down in the new product rule $f=1 - x \implies f'=-1$ and $g=5 x + 8 \implies g'=5$. Popping up a level gives $g'=(5 x + 8)(-1)+(1 - x)(5)$Popping up again (Back to the original problem) gives $f'=(4 - 6 x)(- 10 x - 3)+(\left(1 - x\right) \left(5 x + 8\right))(-6)=5 \left(1 - x\right) \left(4 - 6 x\right) + \left(5 x + 8\right) \left(6 x - 6\right) + \left(5 x + 8\right) \left(6 x - 4\right)$}
\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 = (4 - 6 x)(1 - x)(5 x + 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%284%20-%206%20x%29%281%20-%20x%29%285%20x%20%2B%208%29%20" alt="LaTeX: \displaystyle y = (4 - 6 x)(1 - x)(5 x + 8) " data-equation-content=" \displaystyle y = (4 - 6 x)(1 - x)(5 x + 8) " /> .</p> </p>
<p> <p>Identifying <img class="equation_image" title=" \displaystyle f=4 - 6 x " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20-%206%20x%20" alt="LaTeX: \displaystyle f=4 - 6 x " data-equation-content=" \displaystyle f=4 - 6 x " /> and <img class="equation_image" title=" \displaystyle g=\left(1 - x\right) \left(5 x + 8\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%281%20-%20x%5Cright%29%20%5Cleft%285%20x%20%2B%208%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(1 - x\right) \left(5 x + 8\right) " data-equation-content=" \displaystyle g=\left(1 - x\right) \left(5 x + 8\right) " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=4 - 6 x \implies f'=-6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20-%206%20x%20%5Cimplies%20f%27%3D-6%20" alt="LaTeX: \displaystyle f=4 - 6 x \implies f'=-6 " data-equation-content=" \displaystyle f=4 - 6 x \implies f'=-6 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(1 - x\right) \left(5 x + 8\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%281%20-%20x%5Cright%29%20%5Cleft%285%20x%20%2B%208%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(1 - x\right) \left(5 x + 8\right) " data-equation-content=" \displaystyle g = \left(1 - x\right) \left(5 x + 8\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=1 - x \implies f'=-1 " src="/equation_images/%20%5Cdisplaystyle%20f%3D1%20-%20x%20%5Cimplies%20f%27%3D-1%20" alt="LaTeX: \displaystyle f=1 - x \implies f'=-1 " data-equation-content=" \displaystyle f=1 - x \implies f'=-1 " /> and <img class="equation_image" title=" \displaystyle g=5 x + 8 \implies g'=5 " src="/equation_images/%20%5Cdisplaystyle%20g%3D5%20x%20%2B%208%20%5Cimplies%20g%27%3D5%20" alt="LaTeX: \displaystyle g=5 x + 8 \implies g'=5 " data-equation-content=" \displaystyle g=5 x + 8 \implies g'=5 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(5 x + 8)(-1)+(1 - x)(5) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%285%20x%20%2B%208%29%28-1%29%2B%281%20-%20x%29%285%29%20" alt="LaTeX: \displaystyle g'=(5 x + 8)(-1)+(1 - x)(5) " data-equation-content=" \displaystyle g'=(5 x + 8)(-1)+(1 - x)(5) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(4 - 6 x)(- 10 x - 3)+(\left(1 - x\right) \left(5 x + 8\right))(-6)=5 \left(1 - x\right) \left(4 - 6 x\right) + \left(5 x + 8\right) \left(6 x - 6\right) + \left(5 x + 8\right) \left(6 x - 4\right) " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%284%20-%206%20x%29%28-%2010%20x%20-%203%29%2B%28%5Cleft%281%20-%20x%5Cright%29%20%5Cleft%285%20x%20%2B%208%5Cright%29%29%28-6%29%3D5%20%5Cleft%281%20-%20x%5Cright%29%20%5Cleft%284%20-%206%20x%5Cright%29%20%2B%20%5Cleft%285%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20-%206%5Cright%29%20%2B%20%5Cleft%285%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20-%204%5Cright%29%20" alt="LaTeX: \displaystyle f'=(4 - 6 x)(- 10 x - 3)+(\left(1 - x\right) \left(5 x + 8\right))(-6)=5 \left(1 - x\right) \left(4 - 6 x\right) + \left(5 x + 8\right) \left(6 x - 6\right) + \left(5 x + 8\right) \left(6 x - 4\right) " data-equation-content=" \displaystyle f'=(4 - 6 x)(- 10 x - 3)+(\left(1 - x\right) \left(5 x + 8\right))(-6)=5 \left(1 - x\right) \left(4 - 6 x\right) + \left(5 x + 8\right) \left(6 x - 6\right) + \left(5 x + 8\right) \left(6 x - 4\right) " /> </p> </p>