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Find the derivative of \(\displaystyle y = (x^{3} - 6 x^{2} - 9 x - 5)(\sin{\left(x \right)})(4 x^{3} - 2 x^{2} - 5 x + 6)\).
Identifying \(\displaystyle f=x^{3} - 6 x^{2} - 9 x - 5\) and \(\displaystyle g=\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)}\) and using the product rule with \(\displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 \implies f'=3 x^{2} - 12 x - 9\). This leaves g as \(\displaystyle g = \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)}\) 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 \(\displaystyle g=4 x^{3} - 2 x^{2} - 5 x + 6 \implies g'=12 x^{2} - 4 x - 5\). Popping up a level gives \(\displaystyle g'=(4 x^{3} - 2 x^{2} - 5 x + 6)(\cos{\left(x \right)})+(\sin{\left(x \right)})(12 x^{2} - 4 x - 5)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(x^{3} - 6 x^{2} - 9 x - 5)(\left(12 x^{2} - 4 x - 5\right) \sin{\left(x \right)} + \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)})+(\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)})(3 x^{2} - 12 x - 9)=\left(3 x^{2} - 12 x - 9\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} + \left(12 x^{2} - 4 x - 5\right) \left(x^{3} - 6 x^{2} - 9 x - 5\right) \sin{\left(x \right)} + \left(x^{3} - 6 x^{2} - 9 x - 5\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)}\)
\begin{question}Find the derivative of $y = (x^{3} - 6 x^{2} - 9 x - 5)(\sin{\left(x \right)})(4 x^{3} - 2 x^{2} - 5 x + 6)$.
\soln{9cm}{Identifying $f=x^{3} - 6 x^{2} - 9 x - 5$ and $g=\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)}$ and using the product rule with $f=x^{3} - 6 x^{2} - 9 x - 5 \implies f'=3 x^{2} - 12 x - 9$. This leaves g as $g = \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)}$ 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 $g=4 x^{3} - 2 x^{2} - 5 x + 6 \implies g'=12 x^{2} - 4 x - 5$. Popping up a level gives $g'=(4 x^{3} - 2 x^{2} - 5 x + 6)(\cos{\left(x \right)})+(\sin{\left(x \right)})(12 x^{2} - 4 x - 5)$Popping up again (Back to the original problem) gives $f'=(x^{3} - 6 x^{2} - 9 x - 5)(\left(12 x^{2} - 4 x - 5\right) \sin{\left(x \right)} + \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)})+(\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)})(3 x^{2} - 12 x - 9)=\left(3 x^{2} - 12 x - 9\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} + \left(12 x^{2} - 4 x - 5\right) \left(x^{3} - 6 x^{2} - 9 x - 5\right) \sin{\left(x \right)} + \left(x^{3} - 6 x^{2} - 9 x - 5\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)}$}
\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 = (x^{3} - 6 x^{2} - 9 x - 5)(\sin{\left(x \right)})(4 x^{3} - 2 x^{2} - 5 x + 6) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%29%20" alt="LaTeX: \displaystyle y = (x^{3} - 6 x^{2} - 9 x - 5)(\sin{\left(x \right)})(4 x^{3} - 2 x^{2} - 5 x + 6) " data-equation-content=" \displaystyle y = (x^{3} - 6 x^{2} - 9 x - 5)(\sin{\left(x \right)})(4 x^{3} - 2 x^{2} - 5 x + 6) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%20" alt="LaTeX: \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 " data-equation-content=" \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 " /> and <img class="equation_image" title=" \displaystyle g=\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 \implies f'=3 x^{2} - 12 x - 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%20%5Cimplies%20f%27%3D3%20x%5E%7B2%7D%20-%2012%20x%20-%209%20" alt="LaTeX: \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 \implies f'=3 x^{2} - 12 x - 9 " data-equation-content=" \displaystyle f=x^{3} - 6 x^{2} - 9 x - 5 \implies f'=3 x^{2} - 12 x - 9 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g = \left(4 x^{3} - 2 x^{2} - 5 x + 6\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=4 x^{3} - 2 x^{2} - 5 x + 6 \implies g'=12 x^{2} - 4 x - 5 " src="/equation_images/%20%5Cdisplaystyle%20g%3D4%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%20%5Cimplies%20g%27%3D12%20x%5E%7B2%7D%20-%204%20x%20-%205%20" alt="LaTeX: \displaystyle g=4 x^{3} - 2 x^{2} - 5 x + 6 \implies g'=12 x^{2} - 4 x - 5 " data-equation-content=" \displaystyle g=4 x^{3} - 2 x^{2} - 5 x + 6 \implies g'=12 x^{2} - 4 x - 5 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(4 x^{3} - 2 x^{2} - 5 x + 6)(\cos{\left(x \right)})+(\sin{\left(x \right)})(12 x^{2} - 4 x - 5) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2812%20x%5E%7B2%7D%20-%204%20x%20-%205%29%20" alt="LaTeX: \displaystyle g'=(4 x^{3} - 2 x^{2} - 5 x + 6)(\cos{\left(x \right)})+(\sin{\left(x \right)})(12 x^{2} - 4 x - 5) " data-equation-content=" \displaystyle g'=(4 x^{3} - 2 x^{2} - 5 x + 6)(\cos{\left(x \right)})+(\sin{\left(x \right)})(12 x^{2} - 4 x - 5) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(x^{3} - 6 x^{2} - 9 x - 5)(\left(12 x^{2} - 4 x - 5\right) \sin{\left(x \right)} + \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)})+(\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)})(3 x^{2} - 12 x - 9)=\left(3 x^{2} - 12 x - 9\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} + \left(12 x^{2} - 4 x - 5\right) \left(x^{3} - 6 x^{2} - 9 x - 5\right) \sin{\left(x \right)} + \left(x^{3} - 6 x^{2} - 9 x - 5\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%29%28%5Cleft%2812%20x%5E%7B2%7D%20-%204%20x%20-%205%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%283%20x%5E%7B2%7D%20-%2012%20x%20-%209%29%3D%5Cleft%283%20x%5E%7B2%7D%20-%2012%20x%20-%209%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%5E%7B2%7D%20-%204%20x%20-%205%5Cright%29%20%5Cleft%28x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%209%20x%20-%205%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%206%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(x^{3} - 6 x^{2} - 9 x - 5)(\left(12 x^{2} - 4 x - 5\right) \sin{\left(x \right)} + \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)})+(\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)})(3 x^{2} - 12 x - 9)=\left(3 x^{2} - 12 x - 9\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} + \left(12 x^{2} - 4 x - 5\right) \left(x^{3} - 6 x^{2} - 9 x - 5\right) \sin{\left(x \right)} + \left(x^{3} - 6 x^{2} - 9 x - 5\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(x^{3} - 6 x^{2} - 9 x - 5)(\left(12 x^{2} - 4 x - 5\right) \sin{\left(x \right)} + \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)})+(\left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)})(3 x^{2} - 12 x - 9)=\left(3 x^{2} - 12 x - 9\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \sin{\left(x \right)} + \left(12 x^{2} - 4 x - 5\right) \left(x^{3} - 6 x^{2} - 9 x - 5\right) \sin{\left(x \right)} + \left(x^{3} - 6 x^{2} - 9 x - 5\right) \left(4 x^{3} - 2 x^{2} - 5 x + 6\right) \cos{\left(x \right)} " /> </p> </p>