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Identifying \(\displaystyle f=\log{\left(x \right)}\) and \(\displaystyle g=\left(2 x^{2} - 2 x + 3\right) e^{x}\) and using the product rule with \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\). This leaves g as \(\displaystyle g = \left(2 x^{2} - 2 x + 3\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=2 x^{2} - 2 x + 3 \implies f'=4 x - 2\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(4 x - 2)+(2 x^{2} - 2 x + 3)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\log{\left(x \right)})(\left(4 x - 2\right) e^{x} + \left(2 x^{2} - 2 x + 3\right) e^{x})+(\left(2 x^{2} - 2 x + 3\right) e^{x})(\frac{1}{x})=\left(4 x - 2\right) e^{x} \log{\left(x \right)} + \left(2 x^{2} - 2 x + 3\right) e^{x} \log{\left(x \right)} + \frac{\left(2 x^{2} - 2 x + 3\right) e^{x}}{x}\)

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\begin{question}Find the derivative of $y = (\log{\left(x \right)})(2 x^{2} - 2 x + 3)(e^{x})$.
    \soln{9cm}{Identifying $f=\log{\left(x \right)}$ and $g=\left(2 x^{2} - 2 x + 3\right) e^{x}$ and using the product rule with $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$. This leaves g as $g = \left(2 x^{2} - 2 x + 3\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=2 x^{2} - 2 x + 3 \implies f'=4 x - 2$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(4 x - 2)+(2 x^{2} - 2 x + 3)(e^{x})$Popping up again (Back to the original problem) gives $f'=(\log{\left(x \right)})(\left(4 x - 2\right) e^{x} + \left(2 x^{2} - 2 x + 3\right) e^{x})+(\left(2 x^{2} - 2 x + 3\right) e^{x})(\frac{1}{x})=\left(4 x - 2\right) e^{x} \log{\left(x \right)} + \left(2 x^{2} - 2 x + 3\right) e^{x} \log{\left(x \right)} + \frac{\left(2 x^{2} - 2 x + 3\right) e^{x}}{x}$}

\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 = (\log{\left(x \right)})(2 x^{2} - 2 x + 3)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle y = (\log{\left(x \right)})(2 x^{2} - 2 x + 3)(e^{x}) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(2 x^{2} - 2 x + 3)(e^{x}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} " data-equation-content=" \displaystyle f=\log{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(2 x^{2} - 2 x + 3\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(2 x^{2} - 2 x + 3\right) e^{x} " data-equation-content=" \displaystyle g=\left(2 x^{2} - 2 x + 3\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " data-equation-content=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(2 x^{2} - 2 x + 3\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(2 x^{2} - 2 x + 3\right) e^{x} " data-equation-content=" \displaystyle g = \left(2 x^{2} - 2 x + 3\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=2 x^{2} - 2 x + 3 \implies f'=4 x - 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D2%20x%5E%7B2%7D%20-%202%20x%20%2B%203%20%5Cimplies%20f%27%3D4%20x%20-%202%20" alt="LaTeX:  \displaystyle f=2 x^{2} - 2 x + 3 \implies f'=4 x - 2 " data-equation-content=" \displaystyle f=2 x^{2} - 2 x + 3 \implies f'=4 x - 2 " />  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})(4 x - 2)+(2 x^{2} - 2 x + 3)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%284%20x%20-%202%29%2B%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle g'=(e^{x})(4 x - 2)+(2 x^{2} - 2 x + 3)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(4 x - 2)+(2 x^{2} - 2 x + 3)(e^{x}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(\left(4 x - 2\right) e^{x} + \left(2 x^{2} - 2 x + 3\right) e^{x})+(\left(2 x^{2} - 2 x + 3\right) e^{x})(\frac{1}{x})=\left(4 x - 2\right) e^{x} \log{\left(x \right)} + \left(2 x^{2} - 2 x + 3\right) e^{x} \log{\left(x \right)} + \frac{\left(2 x^{2} - 2 x + 3\right) e^{x}}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%284%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D%5Cleft%284%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%282%20x%5E%7B2%7D%20-%202%20x%20%2B%203%5Cright%29%20e%5E%7Bx%7D%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(\log{\left(x \right)})(\left(4 x - 2\right) e^{x} + \left(2 x^{2} - 2 x + 3\right) e^{x})+(\left(2 x^{2} - 2 x + 3\right) e^{x})(\frac{1}{x})=\left(4 x - 2\right) e^{x} \log{\left(x \right)} + \left(2 x^{2} - 2 x + 3\right) e^{x} \log{\left(x \right)} + \frac{\left(2 x^{2} - 2 x + 3\right) e^{x}}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(\left(4 x - 2\right) e^{x} + \left(2 x^{2} - 2 x + 3\right) e^{x})+(\left(2 x^{2} - 2 x + 3\right) e^{x})(\frac{1}{x})=\left(4 x - 2\right) e^{x} \log{\left(x \right)} + \left(2 x^{2} - 2 x + 3\right) e^{x} \log{\left(x \right)} + \frac{\left(2 x^{2} - 2 x + 3\right) e^{x}}{x} " /> </p> </p>