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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (7 x - 4)(7 - 5 x)(e^{x})\).
Identifying \(\displaystyle f=7 x - 4\) and \(\displaystyle g=\left(7 - 5 x\right) e^{x}\) and using the product rule with \(\displaystyle f=7 x - 4 \implies f'=7\). This leaves g as \(\displaystyle g = \left(7 - 5 x\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=7 - 5 x \implies f'=-5\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(-5)+(7 - 5 x)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(7 x - 4)(\left(7 - 5 x\right) e^{x} - 5 e^{x})+(\left(7 - 5 x\right) e^{x})(7)=\left(7 - 5 x\right) \left(7 x - 4\right) e^{x} + \left(20 - 35 x\right) e^{x} + \left(49 - 35 x\right) e^{x}\)
\begin{question}Find the derivative of $y = (7 x - 4)(7 - 5 x)(e^{x})$.
\soln{9cm}{Identifying $f=7 x - 4$ and $g=\left(7 - 5 x\right) e^{x}$ and using the product rule with $f=7 x - 4 \implies f'=7$. This leaves g as $g = \left(7 - 5 x\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=7 - 5 x \implies f'=-5$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(-5)+(7 - 5 x)(e^{x})$Popping up again (Back to the original problem) gives $f'=(7 x - 4)(\left(7 - 5 x\right) e^{x} - 5 e^{x})+(\left(7 - 5 x\right) e^{x})(7)=\left(7 - 5 x\right) \left(7 x - 4\right) e^{x} + \left(20 - 35 x\right) e^{x} + \left(49 - 35 x\right) e^{x}$}
\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 = (7 x - 4)(7 - 5 x)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%287%20x%20-%204%29%287%20-%205%20x%29%28e%5E%7Bx%7D%29%20" alt="LaTeX: \displaystyle y = (7 x - 4)(7 - 5 x)(e^{x}) " data-equation-content=" \displaystyle y = (7 x - 4)(7 - 5 x)(e^{x}) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=7 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%3D7%20x%20-%204%20" alt="LaTeX: \displaystyle f=7 x - 4 " data-equation-content=" \displaystyle f=7 x - 4 " /> and <img class="equation_image" title=" \displaystyle g=\left(7 - 5 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%287%20-%205%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g=\left(7 - 5 x\right) e^{x} " data-equation-content=" \displaystyle g=\left(7 - 5 x\right) e^{x} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=7 x - 4 \implies f'=7 " src="/equation_images/%20%5Cdisplaystyle%20f%3D7%20x%20-%204%20%5Cimplies%20f%27%3D7%20" alt="LaTeX: \displaystyle f=7 x - 4 \implies f'=7 " data-equation-content=" \displaystyle f=7 x - 4 \implies f'=7 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(7 - 5 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%287%20-%205%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g = \left(7 - 5 x\right) e^{x} " data-equation-content=" \displaystyle g = \left(7 - 5 x\right) e^{x} " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=7 - 5 x \implies f'=-5 " src="/equation_images/%20%5Cdisplaystyle%20f%3D7%20-%205%20x%20%5Cimplies%20f%27%3D-5%20" alt="LaTeX: \displaystyle f=7 - 5 x \implies f'=-5 " data-equation-content=" \displaystyle f=7 - 5 x \implies f'=-5 " /> and <img class="equation_image" title=" \displaystyle g=e^{x} \implies g'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3De%5E%7Bx%7D%20%5Cimplies%20g%27%3De%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g=e^{x} \implies g'=e^{x} " data-equation-content=" \displaystyle g=e^{x} \implies g'=e^{x} " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(e^{x})(-5)+(7 - 5 x)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%28-5%29%2B%287%20-%205%20x%29%28e%5E%7Bx%7D%29%20" alt="LaTeX: \displaystyle g'=(e^{x})(-5)+(7 - 5 x)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(-5)+(7 - 5 x)(e^{x}) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(7 x - 4)(\left(7 - 5 x\right) e^{x} - 5 e^{x})+(\left(7 - 5 x\right) e^{x})(7)=\left(7 - 5 x\right) \left(7 x - 4\right) e^{x} + \left(20 - 35 x\right) e^{x} + \left(49 - 35 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%287%20x%20-%204%29%28%5Cleft%287%20-%205%20x%5Cright%29%20e%5E%7Bx%7D%20-%205%20e%5E%7Bx%7D%29%2B%28%5Cleft%287%20-%205%20x%5Cright%29%20e%5E%7Bx%7D%29%287%29%3D%5Cleft%287%20-%205%20x%5Cright%29%20%5Cleft%287%20x%20-%204%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2820%20-%2035%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2849%20-%2035%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(7 x - 4)(\left(7 - 5 x\right) e^{x} - 5 e^{x})+(\left(7 - 5 x\right) e^{x})(7)=\left(7 - 5 x\right) \left(7 x - 4\right) e^{x} + \left(20 - 35 x\right) e^{x} + \left(49 - 35 x\right) e^{x} " data-equation-content=" \displaystyle f'=(7 x - 4)(\left(7 - 5 x\right) e^{x} - 5 e^{x})+(\left(7 - 5 x\right) e^{x})(7)=\left(7 - 5 x\right) \left(7 x - 4\right) e^{x} + \left(20 - 35 x\right) e^{x} + \left(49 - 35 x\right) e^{x} " /> </p> </p>