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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (e^{x})(5 - 2 x)(3 x + 1)\).
Identifying \(\displaystyle f=e^{x}\) and \(\displaystyle g=\left(5 - 2 x\right) \left(3 x + 1\right)\) and using the product rule with \(\displaystyle f=e^{x} \implies f'=e^{x}\). This leaves g as \(\displaystyle g = \left(5 - 2 x\right) \left(3 x + 1\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=5 - 2 x \implies f'=-2\) and \(\displaystyle g=3 x + 1 \implies g'=3\). Popping up a level gives \(\displaystyle g'=(3 x + 1)(-2)+(5 - 2 x)(3)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(e^{x})(13 - 12 x)+(\left(5 - 2 x\right) \left(3 x + 1\right))(e^{x})=\left(5 - 2 x\right) \left(3 x + 1\right) e^{x} + \left(15 - 6 x\right) e^{x} + \left(- 6 x - 2\right) e^{x}\)
\begin{question}Find the derivative of $y = (e^{x})(5 - 2 x)(3 x + 1)$.
\soln{9cm}{Identifying $f=e^{x}$ and $g=\left(5 - 2 x\right) \left(3 x + 1\right)$ and using the product rule with $f=e^{x} \implies f'=e^{x}$. This leaves g as $g = \left(5 - 2 x\right) \left(3 x + 1\right)$ which also requires the product rule. Pushing down in the new product rule $f=5 - 2 x \implies f'=-2$ and $g=3 x + 1 \implies g'=3$. Popping up a level gives $g'=(3 x + 1)(-2)+(5 - 2 x)(3)$Popping up again (Back to the original problem) gives $f'=(e^{x})(13 - 12 x)+(\left(5 - 2 x\right) \left(3 x + 1\right))(e^{x})=\left(5 - 2 x\right) \left(3 x + 1\right) e^{x} + \left(15 - 6 x\right) e^{x} + \left(- 6 x - 2\right) e^{x}$}
\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 = (e^{x})(5 - 2 x)(3 x + 1) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28e%5E%7Bx%7D%29%285%20-%202%20x%29%283%20x%20%2B%201%29%20" alt="LaTeX: \displaystyle y = (e^{x})(5 - 2 x)(3 x + 1) " data-equation-content=" \displaystyle y = (e^{x})(5 - 2 x)(3 x + 1) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f=e^{x} " data-equation-content=" \displaystyle f=e^{x} " /> and <img class="equation_image" title=" \displaystyle g=\left(5 - 2 x\right) \left(3 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%285%20-%202%20x%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(5 - 2 x\right) \left(3 x + 1\right) " data-equation-content=" \displaystyle g=\left(5 - 2 x\right) \left(3 x + 1\right) " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=e^{x} \implies f'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20%5Cimplies%20f%27%3De%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f=e^{x} \implies f'=e^{x} " data-equation-content=" \displaystyle f=e^{x} \implies f'=e^{x} " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(5 - 2 x\right) \left(3 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%285%20-%202%20x%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(5 - 2 x\right) \left(3 x + 1\right) " data-equation-content=" \displaystyle g = \left(5 - 2 x\right) \left(3 x + 1\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=5 - 2 x \implies f'=-2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20-%202%20x%20%5Cimplies%20f%27%3D-2%20" alt="LaTeX: \displaystyle f=5 - 2 x \implies f'=-2 " data-equation-content=" \displaystyle f=5 - 2 x \implies f'=-2 " /> and <img class="equation_image" title=" \displaystyle g=3 x + 1 \implies g'=3 " src="/equation_images/%20%5Cdisplaystyle%20g%3D3%20x%20%2B%201%20%5Cimplies%20g%27%3D3%20" alt="LaTeX: \displaystyle g=3 x + 1 \implies g'=3 " data-equation-content=" \displaystyle g=3 x + 1 \implies g'=3 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(3 x + 1)(-2)+(5 - 2 x)(3) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%283%20x%20%2B%201%29%28-2%29%2B%285%20-%202%20x%29%283%29%20" alt="LaTeX: \displaystyle g'=(3 x + 1)(-2)+(5 - 2 x)(3) " data-equation-content=" \displaystyle g'=(3 x + 1)(-2)+(5 - 2 x)(3) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(e^{x})(13 - 12 x)+(\left(5 - 2 x\right) \left(3 x + 1\right))(e^{x})=\left(5 - 2 x\right) \left(3 x + 1\right) e^{x} + \left(15 - 6 x\right) e^{x} + \left(- 6 x - 2\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28e%5E%7Bx%7D%29%2813%20-%2012%20x%29%2B%28%5Cleft%285%20-%202%20x%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%29%28e%5E%7Bx%7D%29%3D%5Cleft%285%20-%202%20x%5Cright%29%20%5Cleft%283%20x%20%2B%201%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2815%20-%206%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%206%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(e^{x})(13 - 12 x)+(\left(5 - 2 x\right) \left(3 x + 1\right))(e^{x})=\left(5 - 2 x\right) \left(3 x + 1\right) e^{x} + \left(15 - 6 x\right) e^{x} + \left(- 6 x - 2\right) e^{x} " data-equation-content=" \displaystyle f'=(e^{x})(13 - 12 x)+(\left(5 - 2 x\right) \left(3 x + 1\right))(e^{x})=\left(5 - 2 x\right) \left(3 x + 1\right) e^{x} + \left(15 - 6 x\right) e^{x} + \left(- 6 x - 2\right) e^{x} " /> </p> </p>