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Identifying \(\displaystyle f=3 x - 8\) and \(\displaystyle g=\left(x - 1\right) e^{x}\) and using the product rule with \(\displaystyle f=3 x - 8 \implies f'=3\). This leaves g as \(\displaystyle g = \left(x - 1\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=x - 1 \implies f'=1\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(1)+(x - 1)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(3 x - 8)(\left(x - 1\right) e^{x} + e^{x})+(\left(x - 1\right) e^{x})(3)=\left(x - 1\right) \left(3 x - 8\right) e^{x} + \left(3 x - 8\right) e^{x} + \left(3 x - 3\right) e^{x}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (3 x - 8)(x - 1)(e^{x})$.
    \soln{9cm}{Identifying $f=3 x - 8$ and $g=\left(x - 1\right) e^{x}$ and using the product rule with $f=3 x - 8 \implies f'=3$. This leaves g as $g = \left(x - 1\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=x - 1 \implies f'=1$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(1)+(x - 1)(e^{x})$Popping up again (Back to the original problem) gives $f'=(3 x - 8)(\left(x - 1\right) e^{x} + e^{x})+(\left(x - 1\right) e^{x})(3)=\left(x - 1\right) \left(3 x - 8\right) e^{x} + \left(3 x - 8\right) e^{x} + \left(3 x - 3\right) e^{x}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (3 x - 8)(x - 1)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%283%20x%20-%208%29%28x%20-%201%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle y = (3 x - 8)(x - 1)(e^{x}) " data-equation-content=" \displaystyle y = (3 x - 8)(x - 1)(e^{x}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=3 x - 8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D3%20x%20-%208%20" alt="LaTeX:  \displaystyle f=3 x - 8 " data-equation-content=" \displaystyle f=3 x - 8 " />  and  <img class="equation_image" title=" \displaystyle g=\left(x - 1\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28x%20-%201%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(x - 1\right) e^{x} " data-equation-content=" \displaystyle g=\left(x - 1\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=3 x - 8 \implies f'=3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D3%20x%20-%208%20%5Cimplies%20f%27%3D3%20" alt="LaTeX:  \displaystyle f=3 x - 8 \implies f'=3 " data-equation-content=" \displaystyle f=3 x - 8 \implies f'=3 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(x - 1\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28x%20-%201%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(x - 1\right) e^{x} " data-equation-content=" \displaystyle g = \left(x - 1\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=x - 1 \implies f'=1 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%20-%201%20%5Cimplies%20f%27%3D1%20" alt="LaTeX:  \displaystyle f=x - 1 \implies f'=1 " data-equation-content=" \displaystyle f=x - 1 \implies f'=1 " />  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})(1)+(x - 1)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%281%29%2B%28x%20-%201%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle g'=(e^{x})(1)+(x - 1)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(1)+(x - 1)(e^{x}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(3 x - 8)(\left(x - 1\right) e^{x} + e^{x})+(\left(x - 1\right) e^{x})(3)=\left(x - 1\right) \left(3 x - 8\right) e^{x} + \left(3 x - 8\right) e^{x} + \left(3 x - 3\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%283%20x%20-%208%29%28%5Cleft%28x%20-%201%5Cright%29%20e%5E%7Bx%7D%20%2B%20e%5E%7Bx%7D%29%2B%28%5Cleft%28x%20-%201%5Cright%29%20e%5E%7Bx%7D%29%283%29%3D%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%283%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%283%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%283%20x%20-%203%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(3 x - 8)(\left(x - 1\right) e^{x} + e^{x})+(\left(x - 1\right) e^{x})(3)=\left(x - 1\right) \left(3 x - 8\right) e^{x} + \left(3 x - 8\right) e^{x} + \left(3 x - 3\right) e^{x} " data-equation-content=" \displaystyle f'=(3 x - 8)(\left(x - 1\right) e^{x} + e^{x})+(\left(x - 1\right) e^{x})(3)=\left(x - 1\right) \left(3 x - 8\right) e^{x} + \left(3 x - 8\right) e^{x} + \left(3 x - 3\right) e^{x} " /> </p> </p>