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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (e^{x})(2 - 5 x)(2 - 7 x)\).
Identifying \(\displaystyle f=e^{x}\) and \(\displaystyle g=\left(2 - 7 x\right) \left(2 - 5 x\right)\) and using the product rule with \(\displaystyle f=e^{x} \implies f'=e^{x}\). This leaves g as \(\displaystyle g = \left(2 - 7 x\right) \left(2 - 5 x\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=2 - 5 x \implies f'=-5\) and \(\displaystyle g=2 - 7 x \implies g'=-7\). Popping up a level gives \(\displaystyle g'=(2 - 7 x)(-5)+(2 - 5 x)(-7)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(e^{x})(70 x - 24)+(\left(2 - 7 x\right) \left(2 - 5 x\right))(e^{x})=\left(2 - 7 x\right) \left(2 - 5 x\right) e^{x} + \left(35 x - 14\right) e^{x} + \left(35 x - 10\right) e^{x}\)
\begin{question}Find the derivative of $y = (e^{x})(2 - 5 x)(2 - 7 x)$.
\soln{9cm}{Identifying $f=e^{x}$ and $g=\left(2 - 7 x\right) \left(2 - 5 x\right)$ and using the product rule with $f=e^{x} \implies f'=e^{x}$. This leaves g as $g = \left(2 - 7 x\right) \left(2 - 5 x\right)$ which also requires the product rule. Pushing down in the new product rule $f=2 - 5 x \implies f'=-5$ and $g=2 - 7 x \implies g'=-7$. Popping up a level gives $g'=(2 - 7 x)(-5)+(2 - 5 x)(-7)$Popping up again (Back to the original problem) gives $f'=(e^{x})(70 x - 24)+(\left(2 - 7 x\right) \left(2 - 5 x\right))(e^{x})=\left(2 - 7 x\right) \left(2 - 5 x\right) e^{x} + \left(35 x - 14\right) e^{x} + \left(35 x - 10\right) e^{x}$}
\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})(2 - 5 x)(2 - 7 x) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28e%5E%7Bx%7D%29%282%20-%205%20x%29%282%20-%207%20x%29%20" alt="LaTeX: \displaystyle y = (e^{x})(2 - 5 x)(2 - 7 x) " data-equation-content=" \displaystyle y = (e^{x})(2 - 5 x)(2 - 7 x) " /> .</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(2 - 7 x\right) \left(2 - 5 x\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20-%207%20x%5Cright%29%20%5Cleft%282%20-%205%20x%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(2 - 7 x\right) \left(2 - 5 x\right) " data-equation-content=" \displaystyle g=\left(2 - 7 x\right) \left(2 - 5 x\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(2 - 7 x\right) \left(2 - 5 x\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20-%207%20x%5Cright%29%20%5Cleft%282%20-%205%20x%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(2 - 7 x\right) \left(2 - 5 x\right) " data-equation-content=" \displaystyle g = \left(2 - 7 x\right) \left(2 - 5 x\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=2 - 5 x \implies f'=-5 " src="/equation_images/%20%5Cdisplaystyle%20f%3D2%20-%205%20x%20%5Cimplies%20f%27%3D-5%20" alt="LaTeX: \displaystyle f=2 - 5 x \implies f'=-5 " data-equation-content=" \displaystyle f=2 - 5 x \implies f'=-5 " /> and <img class="equation_image" title=" \displaystyle g=2 - 7 x \implies g'=-7 " src="/equation_images/%20%5Cdisplaystyle%20g%3D2%20-%207%20x%20%5Cimplies%20g%27%3D-7%20" alt="LaTeX: \displaystyle g=2 - 7 x \implies g'=-7 " data-equation-content=" \displaystyle g=2 - 7 x \implies g'=-7 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(2 - 7 x)(-5)+(2 - 5 x)(-7) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%282%20-%207%20x%29%28-5%29%2B%282%20-%205%20x%29%28-7%29%20" alt="LaTeX: \displaystyle g'=(2 - 7 x)(-5)+(2 - 5 x)(-7) " data-equation-content=" \displaystyle g'=(2 - 7 x)(-5)+(2 - 5 x)(-7) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(e^{x})(70 x - 24)+(\left(2 - 7 x\right) \left(2 - 5 x\right))(e^{x})=\left(2 - 7 x\right) \left(2 - 5 x\right) e^{x} + \left(35 x - 14\right) e^{x} + \left(35 x - 10\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28e%5E%7Bx%7D%29%2870%20x%20-%2024%29%2B%28%5Cleft%282%20-%207%20x%5Cright%29%20%5Cleft%282%20-%205%20x%5Cright%29%29%28e%5E%7Bx%7D%29%3D%5Cleft%282%20-%207%20x%5Cright%29%20%5Cleft%282%20-%205%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2835%20x%20-%2014%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2835%20x%20-%2010%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(e^{x})(70 x - 24)+(\left(2 - 7 x\right) \left(2 - 5 x\right))(e^{x})=\left(2 - 7 x\right) \left(2 - 5 x\right) e^{x} + \left(35 x - 14\right) e^{x} + \left(35 x - 10\right) e^{x} " data-equation-content=" \displaystyle f'=(e^{x})(70 x - 24)+(\left(2 - 7 x\right) \left(2 - 5 x\right))(e^{x})=\left(2 - 7 x\right) \left(2 - 5 x\right) e^{x} + \left(35 x - 14\right) e^{x} + \left(35 x - 10\right) e^{x} " /> </p> </p>