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

Download \(\LaTeX\)

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

\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.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (4 - 3 x)(\sin{\left(x \right)})(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%284%20-%203%20x%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28e%5E%7Bx%7D%29%20" alt="LaTeX:  \displaystyle y = (4 - 3 x)(\sin{\left(x \right)})(e^{x}) " data-equation-content=" \displaystyle y = (4 - 3 x)(\sin{\left(x \right)})(e^{x}) " /> .</p> </p>

HTML for Canvas

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