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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (- 9 x - 9)(- 3 x - 5)(- 2 x - 1)\).
Identifying \(\displaystyle f=- 9 x - 9\) and \(\displaystyle g=\left(- 3 x - 5\right) \left(- 2 x - 1\right)\) and using the product rule with \(\displaystyle f=- 9 x - 9 \implies f'=-9\). This leaves g as \(\displaystyle g = \left(- 3 x - 5\right) \left(- 2 x - 1\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 3 x - 5 \implies f'=-3\) and \(\displaystyle g=- 2 x - 1 \implies g'=-2\). Popping up a level gives \(\displaystyle g'=(- 2 x - 1)(-3)+(- 3 x - 5)(-2)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(- 9 x - 9)(12 x + 13)+(\left(- 3 x - 5\right) \left(- 2 x - 1\right))(-9)=- 3 \left(- 9 x - 9\right) \left(- 2 x - 1\right) + \left(- 3 x - 5\right) \left(18 x + 9\right) + \left(- 3 x - 5\right) \left(18 x + 18\right)\)
\begin{question}Find the derivative of $y = (- 9 x - 9)(- 3 x - 5)(- 2 x - 1)$.
\soln{9cm}{Identifying $f=- 9 x - 9$ and $g=\left(- 3 x - 5\right) \left(- 2 x - 1\right)$ and using the product rule with $f=- 9 x - 9 \implies f'=-9$. This leaves g as $g = \left(- 3 x - 5\right) \left(- 2 x - 1\right)$ which also requires the product rule. Pushing down in the new product rule $f=- 3 x - 5 \implies f'=-3$ and $g=- 2 x - 1 \implies g'=-2$. Popping up a level gives $g'=(- 2 x - 1)(-3)+(- 3 x - 5)(-2)$Popping up again (Back to the original problem) gives $f'=(- 9 x - 9)(12 x + 13)+(\left(- 3 x - 5\right) \left(- 2 x - 1\right))(-9)=- 3 \left(- 9 x - 9\right) \left(- 2 x - 1\right) + \left(- 3 x - 5\right) \left(18 x + 9\right) + \left(- 3 x - 5\right) \left(18 x + 18\right)$}
\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 = (- 9 x - 9)(- 3 x - 5)(- 2 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28-%209%20x%20-%209%29%28-%203%20x%20-%205%29%28-%202%20x%20-%201%29%20" alt="LaTeX: \displaystyle y = (- 9 x - 9)(- 3 x - 5)(- 2 x - 1) " data-equation-content=" \displaystyle y = (- 9 x - 9)(- 3 x - 5)(- 2 x - 1) " /> .</p> </p>
<p> <p>Identifying <img class="equation_image" title=" \displaystyle f=- 9 x - 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%209%20x%20-%209%20" alt="LaTeX: \displaystyle f=- 9 x - 9 " data-equation-content=" \displaystyle f=- 9 x - 9 " /> and <img class="equation_image" title=" \displaystyle g=\left(- 3 x - 5\right) \left(- 2 x - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%203%20x%20-%205%5Cright%29%20%5Cleft%28-%202%20x%20-%201%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(- 3 x - 5\right) \left(- 2 x - 1\right) " data-equation-content=" \displaystyle g=\left(- 3 x - 5\right) \left(- 2 x - 1\right) " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=- 9 x - 9 \implies f'=-9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%209%20x%20-%209%20%5Cimplies%20f%27%3D-9%20" alt="LaTeX: \displaystyle f=- 9 x - 9 \implies f'=-9 " data-equation-content=" \displaystyle f=- 9 x - 9 \implies f'=-9 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- 3 x - 5\right) \left(- 2 x - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%203%20x%20-%205%5Cright%29%20%5Cleft%28-%202%20x%20-%201%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(- 3 x - 5\right) \left(- 2 x - 1\right) " data-equation-content=" \displaystyle g = \left(- 3 x - 5\right) \left(- 2 x - 1\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=- 3 x - 5 \implies f'=-3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%203%20x%20-%205%20%5Cimplies%20f%27%3D-3%20" alt="LaTeX: \displaystyle f=- 3 x - 5 \implies f'=-3 " data-equation-content=" \displaystyle f=- 3 x - 5 \implies f'=-3 " /> and <img class="equation_image" title=" \displaystyle g=- 2 x - 1 \implies g'=-2 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%202%20x%20-%201%20%5Cimplies%20g%27%3D-2%20" alt="LaTeX: \displaystyle g=- 2 x - 1 \implies g'=-2 " data-equation-content=" \displaystyle g=- 2 x - 1 \implies g'=-2 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(- 2 x - 1)(-3)+(- 3 x - 5)(-2) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%202%20x%20-%201%29%28-3%29%2B%28-%203%20x%20-%205%29%28-2%29%20" alt="LaTeX: \displaystyle g'=(- 2 x - 1)(-3)+(- 3 x - 5)(-2) " data-equation-content=" \displaystyle g'=(- 2 x - 1)(-3)+(- 3 x - 5)(-2) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(- 9 x - 9)(12 x + 13)+(\left(- 3 x - 5\right) \left(- 2 x - 1\right))(-9)=- 3 \left(- 9 x - 9\right) \left(- 2 x - 1\right) + \left(- 3 x - 5\right) \left(18 x + 9\right) + \left(- 3 x - 5\right) \left(18 x + 18\right) " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28-%209%20x%20-%209%29%2812%20x%20%2B%2013%29%2B%28%5Cleft%28-%203%20x%20-%205%5Cright%29%20%5Cleft%28-%202%20x%20-%201%5Cright%29%29%28-9%29%3D-%203%20%5Cleft%28-%209%20x%20-%209%5Cright%29%20%5Cleft%28-%202%20x%20-%201%5Cright%29%20%2B%20%5Cleft%28-%203%20x%20-%205%5Cright%29%20%5Cleft%2818%20x%20%2B%209%5Cright%29%20%2B%20%5Cleft%28-%203%20x%20-%205%5Cright%29%20%5Cleft%2818%20x%20%2B%2018%5Cright%29%20" alt="LaTeX: \displaystyle f'=(- 9 x - 9)(12 x + 13)+(\left(- 3 x - 5\right) \left(- 2 x - 1\right))(-9)=- 3 \left(- 9 x - 9\right) \left(- 2 x - 1\right) + \left(- 3 x - 5\right) \left(18 x + 9\right) + \left(- 3 x - 5\right) \left(18 x + 18\right) " data-equation-content=" \displaystyle f'=(- 9 x - 9)(12 x + 13)+(\left(- 3 x - 5\right) \left(- 2 x - 1\right))(-9)=- 3 \left(- 9 x - 9\right) \left(- 2 x - 1\right) + \left(- 3 x - 5\right) \left(18 x + 9\right) + \left(- 3 x - 5\right) \left(18 x + 18\right) " /> </p> </p>