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Find the derivative of \(\displaystyle y = (\sin{\left(x \right)})(4 x - 9)(3 x - 3)\).
Identifying \(\displaystyle f=\sin{\left(x \right)}\) and \(\displaystyle g=\left(3 x - 3\right) \left(4 x - 9\right)\) and using the product rule with \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(3 x - 3\right) \left(4 x - 9\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=4 x - 9 \implies f'=4\) and \(\displaystyle g=3 x - 3 \implies g'=3\). Popping up a level gives \(\displaystyle g'=(3 x - 3)(4)+(4 x - 9)(3)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\sin{\left(x \right)})(24 x - 39)+(\left(3 x - 3\right) \left(4 x - 9\right))(\cos{\left(x \right)})=\left(3 x - 3\right) \left(4 x - 9\right) \cos{\left(x \right)} + \left(12 x - 27\right) \sin{\left(x \right)} + \left(12 x - 12\right) \sin{\left(x \right)}\)
\begin{question}Find the derivative of $y = (\sin{\left(x \right)})(4 x - 9)(3 x - 3)$.
\soln{9cm}{Identifying $f=\sin{\left(x \right)}$ and $g=\left(3 x - 3\right) \left(4 x - 9\right)$ and using the product rule with $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$. This leaves g as $g = \left(3 x - 3\right) \left(4 x - 9\right)$ which also requires the product rule. Pushing down in the new product rule $f=4 x - 9 \implies f'=4$ and $g=3 x - 3 \implies g'=3$. Popping up a level gives $g'=(3 x - 3)(4)+(4 x - 9)(3)$Popping up again (Back to the original problem) gives $f'=(\sin{\left(x \right)})(24 x - 39)+(\left(3 x - 3\right) \left(4 x - 9\right))(\cos{\left(x \right)})=\left(3 x - 3\right) \left(4 x - 9\right) \cos{\left(x \right)} + \left(12 x - 27\right) \sin{\left(x \right)} + \left(12 x - 12\right) \sin{\left(x \right)}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the derivative of <img class="equation_image" title=" \displaystyle y = (\sin{\left(x \right)})(4 x - 9)(3 x - 3) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%284%20x%20-%209%29%283%20x%20-%203%29%20" alt="LaTeX: \displaystyle y = (\sin{\left(x \right)})(4 x - 9)(3 x - 3) " data-equation-content=" \displaystyle y = (\sin{\left(x \right)})(4 x - 9)(3 x - 3) " /> .</p> </p><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(3 x - 3\right) \left(4 x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%283%20x%20-%203%5Cright%29%20%5Cleft%284%20x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(3 x - 3\right) \left(4 x - 9\right) " data-equation-content=" \displaystyle g=\left(3 x - 3\right) \left(4 x - 9\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(3 x - 3\right) \left(4 x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%283%20x%20-%203%5Cright%29%20%5Cleft%284%20x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(3 x - 3\right) \left(4 x - 9\right) " data-equation-content=" \displaystyle g = \left(3 x - 3\right) \left(4 x - 9\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=4 x - 9 \implies f'=4 " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20x%20-%209%20%5Cimplies%20f%27%3D4%20" alt="LaTeX: \displaystyle f=4 x - 9 \implies f'=4 " data-equation-content=" \displaystyle f=4 x - 9 \implies f'=4 " /> and <img class="equation_image" title=" \displaystyle g=3 x - 3 \implies g'=3 " src="/equation_images/%20%5Cdisplaystyle%20g%3D3%20x%20-%203%20%5Cimplies%20g%27%3D3%20" alt="LaTeX: \displaystyle g=3 x - 3 \implies g'=3 " data-equation-content=" \displaystyle g=3 x - 3 \implies g'=3 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(3 x - 3)(4)+(4 x - 9)(3) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%283%20x%20-%203%29%284%29%2B%284%20x%20-%209%29%283%29%20" alt="LaTeX: \displaystyle g'=(3 x - 3)(4)+(4 x - 9)(3) " data-equation-content=" \displaystyle g'=(3 x - 3)(4)+(4 x - 9)(3) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(\sin{\left(x \right)})(24 x - 39)+(\left(3 x - 3\right) \left(4 x - 9\right))(\cos{\left(x \right)})=\left(3 x - 3\right) \left(4 x - 9\right) \cos{\left(x \right)} + \left(12 x - 27\right) \sin{\left(x \right)} + \left(12 x - 12\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2824%20x%20-%2039%29%2B%28%5Cleft%283%20x%20-%203%5Cright%29%20%5Cleft%284%20x%20-%209%5Cright%29%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%283%20x%20-%203%5Cright%29%20%5Cleft%284%20x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%20-%2027%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%20-%2012%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(\sin{\left(x \right)})(24 x - 39)+(\left(3 x - 3\right) \left(4 x - 9\right))(\cos{\left(x \right)})=\left(3 x - 3\right) \left(4 x - 9\right) \cos{\left(x \right)} + \left(12 x - 27\right) \sin{\left(x \right)} + \left(12 x - 12\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(\sin{\left(x \right)})(24 x - 39)+(\left(3 x - 3\right) \left(4 x - 9\right))(\cos{\left(x \right)})=\left(3 x - 3\right) \left(4 x - 9\right) \cos{\left(x \right)} + \left(12 x - 27\right) \sin{\left(x \right)} + \left(12 x - 12\right) \sin{\left(x \right)} " /> </p> </p>