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Identifying \(\displaystyle f=6 x + 2\) and \(\displaystyle g=\left(3 x - 2\right) \cos{\left(x \right)}\) and using the product rule with \(\displaystyle f=6 x + 2 \implies f'=6\). This leaves g as \(\displaystyle g = \left(3 x - 2\right) \cos{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}\) and \(\displaystyle g=3 x - 2 \implies g'=3\). Popping up a level gives \(\displaystyle g'=(3 x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(3)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(6 x + 2)(- \left(3 x - 2\right) \sin{\left(x \right)} + 3 \cos{\left(x \right)})+(\left(3 x - 2\right) \cos{\left(x \right)})(6)=- \left(3 x - 2\right) \left(6 x + 2\right) \sin{\left(x \right)} + \left(18 x - 12\right) \cos{\left(x \right)} + \left(18 x + 6\right) \cos{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (6 x + 2)(\cos{\left(x \right)})(3 x - 2)$.
    \soln{9cm}{Identifying $f=6 x + 2$ and $g=\left(3 x - 2\right) \cos{\left(x \right)}$ and using the product rule with $f=6 x + 2 \implies f'=6$. This leaves g as $g = \left(3 x - 2\right) \cos{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}$ and $g=3 x - 2 \implies g'=3$. Popping up a level gives $g'=(3 x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(3)$Popping up again (Back to the original problem) gives $f'=(6 x + 2)(- \left(3 x - 2\right) \sin{\left(x \right)} + 3 \cos{\left(x \right)})+(\left(3 x - 2\right) \cos{\left(x \right)})(6)=- \left(3 x - 2\right) \left(6 x + 2\right) \sin{\left(x \right)} + \left(18 x - 12\right) \cos{\left(x \right)} + \left(18 x + 6\right) \cos{\left(x \right)}$}

\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 = (6 x + 2)(\cos{\left(x \right)})(3 x - 2) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%286%20x%20%2B%202%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%283%20x%20-%202%29%20" alt="LaTeX:  \displaystyle y = (6 x + 2)(\cos{\left(x \right)})(3 x - 2) " data-equation-content=" \displaystyle y = (6 x + 2)(\cos{\left(x \right)})(3 x - 2) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=6 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D6%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f=6 x + 2 " data-equation-content=" \displaystyle f=6 x + 2 " />  and  <img class="equation_image" title=" \displaystyle g=\left(3 x - 2\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%283%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\left(3 x - 2\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g=\left(3 x - 2\right) \cos{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=6 x + 2 \implies f'=6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D6%20x%20%2B%202%20%5Cimplies%20f%27%3D6%20" alt="LaTeX:  \displaystyle f=6 x + 2 \implies f'=6 " data-equation-content=" \displaystyle f=6 x + 2 \implies f'=6 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(3 x - 2\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%283%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \left(3 x - 2\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g = \left(3 x - 2\right) \cos{\left(x \right)} " />  which also requires the product rule. Pushing down in the new product rule  <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)} " />  and  <img class="equation_image" title=" \displaystyle g=3 x - 2 \implies g'=3 " src="/equation_images/%20%5Cdisplaystyle%20g%3D3%20x%20-%202%20%5Cimplies%20g%27%3D3%20" alt="LaTeX:  \displaystyle g=3 x - 2 \implies g'=3 " data-equation-content=" \displaystyle g=3 x - 2 \implies g'=3 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(3 x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(3) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%283%20x%20-%202%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%283%29%20" alt="LaTeX:  \displaystyle g'=(3 x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(3) " data-equation-content=" \displaystyle g'=(3 x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(3) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(6 x + 2)(- \left(3 x - 2\right) \sin{\left(x \right)} + 3 \cos{\left(x \right)})+(\left(3 x - 2\right) \cos{\left(x \right)})(6)=- \left(3 x - 2\right) \left(6 x + 2\right) \sin{\left(x \right)} + \left(18 x - 12\right) \cos{\left(x \right)} + \left(18 x + 6\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%286%20x%20%2B%202%29%28-%20%5Cleft%283%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%203%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%283%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%286%29%3D-%20%5Cleft%283%20x%20-%202%5Cright%29%20%5Cleft%286%20x%20%2B%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2818%20x%20-%2012%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2818%20x%20%2B%206%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(6 x + 2)(- \left(3 x - 2\right) \sin{\left(x \right)} + 3 \cos{\left(x \right)})+(\left(3 x - 2\right) \cos{\left(x \right)})(6)=- \left(3 x - 2\right) \left(6 x + 2\right) \sin{\left(x \right)} + \left(18 x - 12\right) \cos{\left(x \right)} + \left(18 x + 6\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(6 x + 2)(- \left(3 x - 2\right) \sin{\left(x \right)} + 3 \cos{\left(x \right)})+(\left(3 x - 2\right) \cos{\left(x \right)})(6)=- \left(3 x - 2\right) \left(6 x + 2\right) \sin{\left(x \right)} + \left(18 x - 12\right) \cos{\left(x \right)} + \left(18 x + 6\right) \cos{\left(x \right)} " /> </p> </p>