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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (e^{x})(6 x + 9)(4 x - 1)\).
Identifying \(\displaystyle f=e^{x}\) and \(\displaystyle g=\left(4 x - 1\right) \left(6 x + 9\right)\) and using the product rule with \(\displaystyle f=e^{x} \implies f'=e^{x}\). This leaves g as \(\displaystyle g = \left(4 x - 1\right) \left(6 x + 9\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=6 x + 9 \implies f'=6\) and \(\displaystyle g=4 x - 1 \implies g'=4\). Popping up a level gives \(\displaystyle g'=(4 x - 1)(6)+(6 x + 9)(4)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(e^{x})(48 x + 30)+(\left(4 x - 1\right) \left(6 x + 9\right))(e^{x})=\left(4 x - 1\right) \left(6 x + 9\right) e^{x} + \left(24 x - 6\right) e^{x} + \left(24 x + 36\right) e^{x}\)
\begin{question}Find the derivative of $y = (e^{x})(6 x + 9)(4 x - 1)$.
\soln{9cm}{Identifying $f=e^{x}$ and $g=\left(4 x - 1\right) \left(6 x + 9\right)$ and using the product rule with $f=e^{x} \implies f'=e^{x}$. This leaves g as $g = \left(4 x - 1\right) \left(6 x + 9\right)$ which also requires the product rule. Pushing down in the new product rule $f=6 x + 9 \implies f'=6$ and $g=4 x - 1 \implies g'=4$. Popping up a level gives $g'=(4 x - 1)(6)+(6 x + 9)(4)$Popping up again (Back to the original problem) gives $f'=(e^{x})(48 x + 30)+(\left(4 x - 1\right) \left(6 x + 9\right))(e^{x})=\left(4 x - 1\right) \left(6 x + 9\right) e^{x} + \left(24 x - 6\right) e^{x} + \left(24 x + 36\right) e^{x}$}
\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})(6 x + 9)(4 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28e%5E%7Bx%7D%29%286%20x%20%2B%209%29%284%20x%20-%201%29%20" alt="LaTeX: \displaystyle y = (e^{x})(6 x + 9)(4 x - 1) " data-equation-content=" \displaystyle y = (e^{x})(6 x + 9)(4 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(4 x - 1\right) \left(6 x + 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%284%20x%20-%201%5Cright%29%20%5Cleft%286%20x%20%2B%209%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(4 x - 1\right) \left(6 x + 9\right) " data-equation-content=" \displaystyle g=\left(4 x - 1\right) \left(6 x + 9\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(4 x - 1\right) \left(6 x + 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%284%20x%20-%201%5Cright%29%20%5Cleft%286%20x%20%2B%209%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(4 x - 1\right) \left(6 x + 9\right) " data-equation-content=" \displaystyle g = \left(4 x - 1\right) \left(6 x + 9\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=6 x + 9 \implies f'=6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D6%20x%20%2B%209%20%5Cimplies%20f%27%3D6%20" alt="LaTeX: \displaystyle f=6 x + 9 \implies f'=6 " data-equation-content=" \displaystyle f=6 x + 9 \implies f'=6 " /> and <img class="equation_image" title=" \displaystyle g=4 x - 1 \implies g'=4 " src="/equation_images/%20%5Cdisplaystyle%20g%3D4%20x%20-%201%20%5Cimplies%20g%27%3D4%20" alt="LaTeX: \displaystyle g=4 x - 1 \implies g'=4 " data-equation-content=" \displaystyle g=4 x - 1 \implies g'=4 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(4 x - 1)(6)+(6 x + 9)(4) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%284%20x%20-%201%29%286%29%2B%286%20x%20%2B%209%29%284%29%20" alt="LaTeX: \displaystyle g'=(4 x - 1)(6)+(6 x + 9)(4) " data-equation-content=" \displaystyle g'=(4 x - 1)(6)+(6 x + 9)(4) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(e^{x})(48 x + 30)+(\left(4 x - 1\right) \left(6 x + 9\right))(e^{x})=\left(4 x - 1\right) \left(6 x + 9\right) e^{x} + \left(24 x - 6\right) e^{x} + \left(24 x + 36\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28e%5E%7Bx%7D%29%2848%20x%20%2B%2030%29%2B%28%5Cleft%284%20x%20-%201%5Cright%29%20%5Cleft%286%20x%20%2B%209%5Cright%29%29%28e%5E%7Bx%7D%29%3D%5Cleft%284%20x%20-%201%5Cright%29%20%5Cleft%286%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2824%20x%20-%206%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2824%20x%20%2B%2036%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(e^{x})(48 x + 30)+(\left(4 x - 1\right) \left(6 x + 9\right))(e^{x})=\left(4 x - 1\right) \left(6 x + 9\right) e^{x} + \left(24 x - 6\right) e^{x} + \left(24 x + 36\right) e^{x} " data-equation-content=" \displaystyle f'=(e^{x})(48 x + 30)+(\left(4 x - 1\right) \left(6 x + 9\right))(e^{x})=\left(4 x - 1\right) \left(6 x + 9\right) e^{x} + \left(24 x - 6\right) e^{x} + \left(24 x + 36\right) e^{x} " /> </p> </p>