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Identifying \(\displaystyle f=\cos{\left(x \right)}\) and \(\displaystyle g=\left(5 x^{2} + 2 x - 5\right) e^{x}\) and using the product rule with \(\displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(5 x^{2} + 2 x - 5\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=5 x^{2} + 2 x - 5 \implies f'=10 x + 2\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(10 x + 2)+(5 x^{2} + 2 x - 5)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\cos{\left(x \right)})(\left(10 x + 2\right) e^{x} + \left(5 x^{2} + 2 x - 5\right) e^{x})+(\left(5 x^{2} + 2 x - 5\right) e^{x})(- \sin{\left(x \right)})=\left(10 x + 2\right) e^{x} \cos{\left(x \right)} - \left(5 x^{2} + 2 x - 5\right) e^{x} \sin{\left(x \right)} + \left(5 x^{2} + 2 x - 5\right) e^{x} \cos{\left(x \right)}\)

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\begin{question}Find the derivative of $y = (\cos{\left(x \right)})(5 x^{2} + 2 x - 5)(e^{x})$.
    \soln{9cm}{Identifying $f=\cos{\left(x \right)}$ and $g=\left(5 x^{2} + 2 x - 5\right) e^{x}$ and using the product rule with $f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}$. This leaves g as $g = \left(5 x^{2} + 2 x - 5\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=5 x^{2} + 2 x - 5 \implies f'=10 x + 2$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(10 x + 2)+(5 x^{2} + 2 x - 5)(e^{x})$Popping up again (Back to the original problem) gives $f'=(\cos{\left(x \right)})(\left(10 x + 2\right) e^{x} + \left(5 x^{2} + 2 x - 5\right) e^{x})+(\left(5 x^{2} + 2 x - 5\right) e^{x})(- \sin{\left(x \right)})=\left(10 x + 2\right) e^{x} \cos{\left(x \right)} - \left(5 x^{2} + 2 x - 5\right) e^{x} \sin{\left(x \right)} + \left(5 x^{2} + 2 x - 5\right) e^{x} \cos{\left(x \right)}$}

\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.}

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HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (\cos{\left(x \right)})(5 x^{2} + 2 x - 5)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle y = (\cos{\left(x \right)})(5 x^{2} + 2 x - 5)(e^{x}) " data-equation-content=" \displaystyle y = (\cos{\left(x \right)})(5 x^{2} + 2 x - 5)(e^{x}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\cos{\left(x \right)} " data-equation-content=" \displaystyle f=\cos{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(5 x^{2} + 2 x - 5\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(5 x^{2} + 2 x - 5\right) e^{x} " data-equation-content=" \displaystyle g=\left(5 x^{2} + 2 x - 5\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " data-equation-content=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(5 x^{2} + 2 x - 5\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(5 x^{2} + 2 x - 5\right) e^{x} " data-equation-content=" \displaystyle g = \left(5 x^{2} + 2 x - 5\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=5 x^{2} + 2 x - 5 \implies f'=10 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20x%5E%7B2%7D%20%2B%202%20x%20-%205%20%5Cimplies%20f%27%3D10%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f=5 x^{2} + 2 x - 5 \implies f'=10 x + 2 " data-equation-content=" \displaystyle f=5 x^{2} + 2 x - 5 \implies f'=10 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})(10 x + 2)+(5 x^{2} + 2 x - 5)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%2810%20x%20%2B%202%29%2B%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle g'=(e^{x})(10 x + 2)+(5 x^{2} + 2 x - 5)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(10 x + 2)+(5 x^{2} + 2 x - 5)(e^{x}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\cos{\left(x \right)})(\left(10 x + 2\right) e^{x} + \left(5 x^{2} + 2 x - 5\right) e^{x})+(\left(5 x^{2} + 2 x - 5\right) e^{x})(- \sin{\left(x \right)})=\left(10 x + 2\right) e^{x} \cos{\left(x \right)} - \left(5 x^{2} + 2 x - 5\right) e^{x} \sin{\left(x \right)} + \left(5 x^{2} + 2 x - 5\right) e^{x} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%2810%20x%20%2B%202%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%2810%20x%20%2B%202%5Cright%29%20e%5E%7Bx%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%285%20x%5E%7B2%7D%20%2B%202%20x%20-%205%5Cright%29%20e%5E%7Bx%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(\cos{\left(x \right)})(\left(10 x + 2\right) e^{x} + \left(5 x^{2} + 2 x - 5\right) e^{x})+(\left(5 x^{2} + 2 x - 5\right) e^{x})(- \sin{\left(x \right)})=\left(10 x + 2\right) e^{x} \cos{\left(x \right)} - \left(5 x^{2} + 2 x - 5\right) e^{x} \sin{\left(x \right)} + \left(5 x^{2} + 2 x - 5\right) e^{x} \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(\cos{\left(x \right)})(\left(10 x + 2\right) e^{x} + \left(5 x^{2} + 2 x - 5\right) e^{x})+(\left(5 x^{2} + 2 x - 5\right) e^{x})(- \sin{\left(x \right)})=\left(10 x + 2\right) e^{x} \cos{\left(x \right)} - \left(5 x^{2} + 2 x - 5\right) e^{x} \sin{\left(x \right)} + \left(5 x^{2} + 2 x - 5\right) e^{x} \cos{\left(x \right)} " /> </p> </p>