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Find the derivative of \(\displaystyle y = (8 x^{3} + 4 x^{2} - 8 x + 9)(\cos{\left(x \right)})(- x^{3} + 9 x^{2} + x - 2)\).
Identifying \(\displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9\) and \(\displaystyle g=\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)}\) and using the product rule with \(\displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 \implies f'=24 x^{2} + 8 x - 8\). This leaves g as \(\displaystyle g = \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)}\) 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 \(\displaystyle g=- x^{3} + 9 x^{2} + x - 2 \implies g'=- 3 x^{2} + 18 x + 1\). Popping up a level gives \(\displaystyle g'=(- x^{3} + 9 x^{2} + x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(- 3 x^{2} + 18 x + 1)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(8 x^{3} + 4 x^{2} - 8 x + 9)(\left(- 3 x^{2} + 18 x + 1\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \sin{\left(x \right)})+(\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)})(24 x^{2} + 8 x - 8)=\left(- 3 x^{2} + 18 x + 1\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \cos{\left(x \right)} + \left(24 x^{2} + 8 x - 8\right) \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \sin{\left(x \right)}\)
\begin{question}Find the derivative of $y = (8 x^{3} + 4 x^{2} - 8 x + 9)(\cos{\left(x \right)})(- x^{3} + 9 x^{2} + x - 2)$.
\soln{9cm}{Identifying $f=8 x^{3} + 4 x^{2} - 8 x + 9$ and $g=\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)}$ and using the product rule with $f=8 x^{3} + 4 x^{2} - 8 x + 9 \implies f'=24 x^{2} + 8 x - 8$. This leaves g as $g = \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)}$ 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 $g=- x^{3} + 9 x^{2} + x - 2 \implies g'=- 3 x^{2} + 18 x + 1$. Popping up a level gives $g'=(- x^{3} + 9 x^{2} + x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(- 3 x^{2} + 18 x + 1)$Popping up again (Back to the original problem) gives $f'=(8 x^{3} + 4 x^{2} - 8 x + 9)(\left(- 3 x^{2} + 18 x + 1\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \sin{\left(x \right)})+(\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)})(24 x^{2} + 8 x - 8)=\left(- 3 x^{2} + 18 x + 1\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \cos{\left(x \right)} + \left(24 x^{2} + 8 x - 8\right) \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \sin{\left(x \right)}$}
\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 = (8 x^{3} + 4 x^{2} - 8 x + 9)(\cos{\left(x \right)})(- x^{3} + 9 x^{2} + x - 2) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%288%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%29%20" alt="LaTeX: \displaystyle y = (8 x^{3} + 4 x^{2} - 8 x + 9)(\cos{\left(x \right)})(- x^{3} + 9 x^{2} + x - 2) " data-equation-content=" \displaystyle y = (8 x^{3} + 4 x^{2} - 8 x + 9)(\cos{\left(x \right)})(- x^{3} + 9 x^{2} + x - 2) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%20" alt="LaTeX: \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 " data-equation-content=" \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 " /> and <img class="equation_image" title=" \displaystyle g=\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 \implies f'=24 x^{2} + 8 x - 8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%20%5Cimplies%20f%27%3D24%20x%5E%7B2%7D%20%2B%208%20x%20-%208%20" alt="LaTeX: \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 \implies f'=24 x^{2} + 8 x - 8 " data-equation-content=" \displaystyle f=8 x^{3} + 4 x^{2} - 8 x + 9 \implies f'=24 x^{2} + 8 x - 8 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- x^{3} + 9 x^{2} + 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=- x^{3} + 9 x^{2} + x - 2 \implies g'=- 3 x^{2} + 18 x + 1 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%20%5Cimplies%20g%27%3D-%203%20x%5E%7B2%7D%20%2B%2018%20x%20%2B%201%20" alt="LaTeX: \displaystyle g=- x^{3} + 9 x^{2} + x - 2 \implies g'=- 3 x^{2} + 18 x + 1 " data-equation-content=" \displaystyle g=- x^{3} + 9 x^{2} + x - 2 \implies g'=- 3 x^{2} + 18 x + 1 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(- x^{3} + 9 x^{2} + x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(- 3 x^{2} + 18 x + 1) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%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%28-%203%20x%5E%7B2%7D%20%2B%2018%20x%20%2B%201%29%20" alt="LaTeX: \displaystyle g'=(- x^{3} + 9 x^{2} + x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(- 3 x^{2} + 18 x + 1) " data-equation-content=" \displaystyle g'=(- x^{3} + 9 x^{2} + x - 2)(- \sin{\left(x \right)})+(\cos{\left(x \right)})(- 3 x^{2} + 18 x + 1) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(8 x^{3} + 4 x^{2} - 8 x + 9)(\left(- 3 x^{2} + 18 x + 1\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \sin{\left(x \right)})+(\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)})(24 x^{2} + 8 x - 8)=\left(- 3 x^{2} + 18 x + 1\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \cos{\left(x \right)} + \left(24 x^{2} + 8 x - 8\right) \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%288%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%29%28%5Cleft%28-%203%20x%5E%7B2%7D%20%2B%2018%20x%20%2B%201%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2824%20x%5E%7B2%7D%20%2B%208%20x%20-%208%29%3D%5Cleft%28-%203%20x%5E%7B2%7D%20%2B%2018%20x%20%2B%201%5Cright%29%20%5Cleft%288%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2824%20x%5E%7B2%7D%20%2B%208%20x%20-%208%5Cright%29%20%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%28-%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%20x%20-%202%5Cright%29%20%5Cleft%288%20x%5E%7B3%7D%20%2B%204%20x%5E%7B2%7D%20-%208%20x%20%2B%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(8 x^{3} + 4 x^{2} - 8 x + 9)(\left(- 3 x^{2} + 18 x + 1\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \sin{\left(x \right)})+(\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)})(24 x^{2} + 8 x - 8)=\left(- 3 x^{2} + 18 x + 1\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \cos{\left(x \right)} + \left(24 x^{2} + 8 x - 8\right) \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(8 x^{3} + 4 x^{2} - 8 x + 9)(\left(- 3 x^{2} + 18 x + 1\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \sin{\left(x \right)})+(\left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)})(24 x^{2} + 8 x - 8)=\left(- 3 x^{2} + 18 x + 1\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \cos{\left(x \right)} + \left(24 x^{2} + 8 x - 8\right) \left(- x^{3} + 9 x^{2} + x - 2\right) \cos{\left(x \right)} - \left(- x^{3} + 9 x^{2} + x - 2\right) \left(8 x^{3} + 4 x^{2} - 8 x + 9\right) \sin{\left(x \right)} " /> </p> </p>