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Identifying \(\displaystyle f=8 - 9 x\) and \(\displaystyle g=\left(x - 9\right) \sin{\left(x \right)}\) and using the product rule with \(\displaystyle f=8 - 9 x \implies f'=-9\). This leaves g as \(\displaystyle g = \left(x - 9\right) \sin{\left(x \right)}\) 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 \(\displaystyle g=x - 9 \implies g'=1\). Popping up a level gives \(\displaystyle g'=(x - 9)(\cos{\left(x \right)})+(\sin{\left(x \right)})(1)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(8 - 9 x)(\left(x - 9\right) \cos{\left(x \right)} + \sin{\left(x \right)})+(\left(x - 9\right) \sin{\left(x \right)})(-9)=\left(8 - 9 x\right) \left(x - 9\right) \cos{\left(x \right)} + \left(8 - 9 x\right) \sin{\left(x \right)} + \left(81 - 9 x\right) \sin{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (8 - 9 x)(\sin{\left(x \right)})(x - 9)$.
    \soln{9cm}{Identifying $f=8 - 9 x$ and $g=\left(x - 9\right) \sin{\left(x \right)}$ and using the product rule with $f=8 - 9 x \implies f'=-9$. This leaves g as $g = \left(x - 9\right) \sin{\left(x \right)}$ 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 $g=x - 9 \implies g'=1$. Popping up a level gives $g'=(x - 9)(\cos{\left(x \right)})+(\sin{\left(x \right)})(1)$Popping up again (Back to the original problem) gives $f'=(8 - 9 x)(\left(x - 9\right) \cos{\left(x \right)} + \sin{\left(x \right)})+(\left(x - 9\right) \sin{\left(x \right)})(-9)=\left(8 - 9 x\right) \left(x - 9\right) \cos{\left(x \right)} + \left(8 - 9 x\right) \sin{\left(x \right)} + \left(81 - 9 x\right) \sin{\left(x \right)}$}

\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 = (8 - 9 x)(\sin{\left(x \right)})(x - 9) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%288%20-%209%20x%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28x%20-%209%29%20" alt="LaTeX:  \displaystyle y = (8 - 9 x)(\sin{\left(x \right)})(x - 9) " data-equation-content=" \displaystyle y = (8 - 9 x)(\sin{\left(x \right)})(x - 9) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=8 - 9 x " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20-%209%20x%20" alt="LaTeX:  \displaystyle f=8 - 9 x " data-equation-content=" \displaystyle f=8 - 9 x " />  and  <img class="equation_image" title=" \displaystyle g=\left(x - 9\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(x - 9\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\left(x - 9\right) \sin{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=8 - 9 x \implies f'=-9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20-%209%20x%20%5Cimplies%20f%27%3D-9%20" alt="LaTeX:  \displaystyle f=8 - 9 x \implies f'=-9 " data-equation-content=" \displaystyle f=8 - 9 x \implies f'=-9 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(x - 9\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(x - 9\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g = \left(x - 9\right) \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=x - 9 \implies g'=1 " src="/equation_images/%20%5Cdisplaystyle%20g%3Dx%20-%209%20%5Cimplies%20g%27%3D1%20" alt="LaTeX:  \displaystyle g=x - 9 \implies g'=1 " data-equation-content=" \displaystyle g=x - 9 \implies g'=1 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(x - 9)(\cos{\left(x \right)})+(\sin{\left(x \right)})(1) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28x%20-%209%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%281%29%20" alt="LaTeX:  \displaystyle g'=(x - 9)(\cos{\left(x \right)})+(\sin{\left(x \right)})(1) " data-equation-content=" \displaystyle g'=(x - 9)(\cos{\left(x \right)})+(\sin{\left(x \right)})(1) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(8 - 9 x)(\left(x - 9\right) \cos{\left(x \right)} + \sin{\left(x \right)})+(\left(x - 9\right) \sin{\left(x \right)})(-9)=\left(8 - 9 x\right) \left(x - 9\right) \cos{\left(x \right)} + \left(8 - 9 x\right) \sin{\left(x \right)} + \left(81 - 9 x\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%288%20-%209%20x%29%28%5Cleft%28x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%28x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-9%29%3D%5Cleft%288%20-%209%20x%5Cright%29%20%5Cleft%28x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%288%20-%209%20x%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2881%20-%209%20x%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(8 - 9 x)(\left(x - 9\right) \cos{\left(x \right)} + \sin{\left(x \right)})+(\left(x - 9\right) \sin{\left(x \right)})(-9)=\left(8 - 9 x\right) \left(x - 9\right) \cos{\left(x \right)} + \left(8 - 9 x\right) \sin{\left(x \right)} + \left(81 - 9 x\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(8 - 9 x)(\left(x - 9\right) \cos{\left(x \right)} + \sin{\left(x \right)})+(\left(x - 9\right) \sin{\left(x \right)})(-9)=\left(8 - 9 x\right) \left(x - 9\right) \cos{\left(x \right)} + \left(8 - 9 x\right) \sin{\left(x \right)} + \left(81 - 9 x\right) \sin{\left(x \right)} " /> </p> </p>