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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (3 - 8 x)(- 5 x - 7)(6 x - 4)\).
Identifying \(\displaystyle f=3 - 8 x\) and \(\displaystyle g=\left(- 5 x - 7\right) \left(6 x - 4\right)\) and using the product rule with \(\displaystyle f=3 - 8 x \implies f'=-8\). This leaves g as \(\displaystyle g = \left(- 5 x - 7\right) \left(6 x - 4\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 5 x - 7 \implies f'=-5\) and \(\displaystyle g=6 x - 4 \implies g'=6\). Popping up a level gives \(\displaystyle g'=(6 x - 4)(-5)+(- 5 x - 7)(6)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(3 - 8 x)(- 60 x - 22)+(\left(- 5 x - 7\right) \left(6 x - 4\right))(-8)=\left(3 - 8 x\right) \left(20 - 30 x\right) + \left(3 - 8 x\right) \left(- 30 x - 42\right) - 8 \left(- 5 x - 7\right) \left(6 x - 4\right)\)
\begin{question}Find the derivative of $y = (3 - 8 x)(- 5 x - 7)(6 x - 4)$.
\soln{9cm}{Identifying $f=3 - 8 x$ and $g=\left(- 5 x - 7\right) \left(6 x - 4\right)$ and using the product rule with $f=3 - 8 x \implies f'=-8$. This leaves g as $g = \left(- 5 x - 7\right) \left(6 x - 4\right)$ which also requires the product rule. Pushing down in the new product rule $f=- 5 x - 7 \implies f'=-5$ and $g=6 x - 4 \implies g'=6$. Popping up a level gives $g'=(6 x - 4)(-5)+(- 5 x - 7)(6)$Popping up again (Back to the original problem) gives $f'=(3 - 8 x)(- 60 x - 22)+(\left(- 5 x - 7\right) \left(6 x - 4\right))(-8)=\left(3 - 8 x\right) \left(20 - 30 x\right) + \left(3 - 8 x\right) \left(- 30 x - 42\right) - 8 \left(- 5 x - 7\right) \left(6 x - 4\right)$}
\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 = (3 - 8 x)(- 5 x - 7)(6 x - 4) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%283%20-%208%20x%29%28-%205%20x%20-%207%29%286%20x%20-%204%29%20" alt="LaTeX: \displaystyle y = (3 - 8 x)(- 5 x - 7)(6 x - 4) " data-equation-content=" \displaystyle y = (3 - 8 x)(- 5 x - 7)(6 x - 4) " /> .</p> </p>
<p> <p>Identifying <img class="equation_image" title=" \displaystyle f=3 - 8 x " src="/equation_images/%20%5Cdisplaystyle%20f%3D3%20-%208%20x%20" alt="LaTeX: \displaystyle f=3 - 8 x " data-equation-content=" \displaystyle f=3 - 8 x " /> and <img class="equation_image" title=" \displaystyle g=\left(- 5 x - 7\right) \left(6 x - 4\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%205%20x%20-%207%5Cright%29%20%5Cleft%286%20x%20-%204%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(- 5 x - 7\right) \left(6 x - 4\right) " data-equation-content=" \displaystyle g=\left(- 5 x - 7\right) \left(6 x - 4\right) " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=3 - 8 x \implies f'=-8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D3%20-%208%20x%20%5Cimplies%20f%27%3D-8%20" alt="LaTeX: \displaystyle f=3 - 8 x \implies f'=-8 " data-equation-content=" \displaystyle f=3 - 8 x \implies f'=-8 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- 5 x - 7\right) \left(6 x - 4\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%205%20x%20-%207%5Cright%29%20%5Cleft%286%20x%20-%204%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(- 5 x - 7\right) \left(6 x - 4\right) " data-equation-content=" \displaystyle g = \left(- 5 x - 7\right) \left(6 x - 4\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=- 5 x - 7 \implies f'=-5 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%205%20x%20-%207%20%5Cimplies%20f%27%3D-5%20" alt="LaTeX: \displaystyle f=- 5 x - 7 \implies f'=-5 " data-equation-content=" \displaystyle f=- 5 x - 7 \implies f'=-5 " /> and <img class="equation_image" title=" \displaystyle g=6 x - 4 \implies g'=6 " src="/equation_images/%20%5Cdisplaystyle%20g%3D6%20x%20-%204%20%5Cimplies%20g%27%3D6%20" alt="LaTeX: \displaystyle g=6 x - 4 \implies g'=6 " data-equation-content=" \displaystyle g=6 x - 4 \implies g'=6 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(6 x - 4)(-5)+(- 5 x - 7)(6) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%286%20x%20-%204%29%28-5%29%2B%28-%205%20x%20-%207%29%286%29%20" alt="LaTeX: \displaystyle g'=(6 x - 4)(-5)+(- 5 x - 7)(6) " data-equation-content=" \displaystyle g'=(6 x - 4)(-5)+(- 5 x - 7)(6) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(3 - 8 x)(- 60 x - 22)+(\left(- 5 x - 7\right) \left(6 x - 4\right))(-8)=\left(3 - 8 x\right) \left(20 - 30 x\right) + \left(3 - 8 x\right) \left(- 30 x - 42\right) - 8 \left(- 5 x - 7\right) \left(6 x - 4\right) " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%283%20-%208%20x%29%28-%2060%20x%20-%2022%29%2B%28%5Cleft%28-%205%20x%20-%207%5Cright%29%20%5Cleft%286%20x%20-%204%5Cright%29%29%28-8%29%3D%5Cleft%283%20-%208%20x%5Cright%29%20%5Cleft%2820%20-%2030%20x%5Cright%29%20%2B%20%5Cleft%283%20-%208%20x%5Cright%29%20%5Cleft%28-%2030%20x%20-%2042%5Cright%29%20-%208%20%5Cleft%28-%205%20x%20-%207%5Cright%29%20%5Cleft%286%20x%20-%204%5Cright%29%20" alt="LaTeX: \displaystyle f'=(3 - 8 x)(- 60 x - 22)+(\left(- 5 x - 7\right) \left(6 x - 4\right))(-8)=\left(3 - 8 x\right) \left(20 - 30 x\right) + \left(3 - 8 x\right) \left(- 30 x - 42\right) - 8 \left(- 5 x - 7\right) \left(6 x - 4\right) " data-equation-content=" \displaystyle f'=(3 - 8 x)(- 60 x - 22)+(\left(- 5 x - 7\right) \left(6 x - 4\right))(-8)=\left(3 - 8 x\right) \left(20 - 30 x\right) + \left(3 - 8 x\right) \left(- 30 x - 42\right) - 8 \left(- 5 x - 7\right) \left(6 x - 4\right) " /> </p> </p>