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Identifying \(\displaystyle f=\sin{\left(x \right)}\) and \(\displaystyle g=\left(9 - 3 x\right) \left(x + 7\right)\) and using the product rule with \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(9 - 3 x\right) \left(x + 7\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=x + 7 \implies f'=1\) and \(\displaystyle g=9 - 3 x \implies g'=-3\). Popping up a level gives \(\displaystyle g'=(9 - 3 x)(1)+(x + 7)(-3)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\sin{\left(x \right)})(- 6 x - 12)+(\left(9 - 3 x\right) \left(x + 7\right))(\cos{\left(x \right)})=\left(9 - 3 x\right) \left(x + 7\right) \cos{\left(x \right)} + \left(9 - 3 x\right) \sin{\left(x \right)} + \left(- 3 x - 21\right) \sin{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (\sin{\left(x \right)})(x + 7)(9 - 3 x)$.
    \soln{9cm}{Identifying $f=\sin{\left(x \right)}$ and $g=\left(9 - 3 x\right) \left(x + 7\right)$ and using the product rule with $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$. This leaves g as $g = \left(9 - 3 x\right) \left(x + 7\right)$ which also requires the product rule. Pushing down in the new product rule $f=x + 7 \implies f'=1$ and $g=9 - 3 x \implies g'=-3$. Popping up a level gives $g'=(9 - 3 x)(1)+(x + 7)(-3)$Popping up again (Back to the original problem) gives $f'=(\sin{\left(x \right)})(- 6 x - 12)+(\left(9 - 3 x\right) \left(x + 7\right))(\cos{\left(x \right)})=\left(9 - 3 x\right) \left(x + 7\right) \cos{\left(x \right)} + \left(9 - 3 x\right) \sin{\left(x \right)} + \left(- 3 x - 21\right) \sin{\left(x \right)}$}

\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 = (\sin{\left(x \right)})(x + 7)(9 - 3 x) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28x%20%2B%207%29%289%20-%203%20x%29%20" alt="LaTeX:  \displaystyle y = (\sin{\left(x \right)})(x + 7)(9 - 3 x) " data-equation-content=" \displaystyle y = (\sin{\left(x \right)})(x + 7)(9 - 3 x) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\sin{\left(x \right)} " data-equation-content=" \displaystyle f=\sin{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(9 - 3 x\right) \left(x + 7\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%289%20-%203%20x%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(9 - 3 x\right) \left(x + 7\right) " data-equation-content=" \displaystyle g=\left(9 - 3 x\right) \left(x + 7\right) " />  and using the product rule with  <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)} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(9 - 3 x\right) \left(x + 7\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%289%20-%203%20x%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(9 - 3 x\right) \left(x + 7\right) " data-equation-content=" \displaystyle g = \left(9 - 3 x\right) \left(x + 7\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=x + 7 \implies f'=1 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%20%2B%207%20%5Cimplies%20f%27%3D1%20" alt="LaTeX:  \displaystyle f=x + 7 \implies f'=1 " data-equation-content=" \displaystyle f=x + 7 \implies f'=1 " />  and  <img class="equation_image" title=" \displaystyle g=9 - 3 x \implies g'=-3 " src="/equation_images/%20%5Cdisplaystyle%20g%3D9%20-%203%20x%20%5Cimplies%20g%27%3D-3%20" alt="LaTeX:  \displaystyle g=9 - 3 x \implies g'=-3 " data-equation-content=" \displaystyle g=9 - 3 x \implies g'=-3 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(9 - 3 x)(1)+(x + 7)(-3) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%289%20-%203%20x%29%281%29%2B%28x%20%2B%207%29%28-3%29%20" alt="LaTeX:  \displaystyle g'=(9 - 3 x)(1)+(x + 7)(-3) " data-equation-content=" \displaystyle g'=(9 - 3 x)(1)+(x + 7)(-3) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\sin{\left(x \right)})(- 6 x - 12)+(\left(9 - 3 x\right) \left(x + 7\right))(\cos{\left(x \right)})=\left(9 - 3 x\right) \left(x + 7\right) \cos{\left(x \right)} + \left(9 - 3 x\right) \sin{\left(x \right)} + \left(- 3 x - 21\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%206%20x%20-%2012%29%2B%28%5Cleft%289%20-%203%20x%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%289%20-%203%20x%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%289%20-%203%20x%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%203%20x%20-%2021%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(\sin{\left(x \right)})(- 6 x - 12)+(\left(9 - 3 x\right) \left(x + 7\right))(\cos{\left(x \right)})=\left(9 - 3 x\right) \left(x + 7\right) \cos{\left(x \right)} + \left(9 - 3 x\right) \sin{\left(x \right)} + \left(- 3 x - 21\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(\sin{\left(x \right)})(- 6 x - 12)+(\left(9 - 3 x\right) \left(x + 7\right))(\cos{\left(x \right)})=\left(9 - 3 x\right) \left(x + 7\right) \cos{\left(x \right)} + \left(9 - 3 x\right) \sin{\left(x \right)} + \left(- 3 x - 21\right) \sin{\left(x \right)} " /> </p> </p>