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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (4 x^{2} - 2 x + 2)(\log{\left(x \right)})(6 x^{2} + 9 x - 9)\).
Identifying \(\displaystyle f=4 x^{2} - 2 x + 2\) and \(\displaystyle g=\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)}\) and using the product rule with \(\displaystyle f=4 x^{2} - 2 x + 2 \implies f'=8 x - 2\). This leaves g as \(\displaystyle g = \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\) and \(\displaystyle g=6 x^{2} + 9 x - 9 \implies g'=12 x + 9\). Popping up a level gives \(\displaystyle g'=(6 x^{2} + 9 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(12 x + 9)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(4 x^{2} - 2 x + 2)(\left(12 x + 9\right) \log{\left(x \right)} + \frac{6 x^{2} + 9 x - 9}{x})+(\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)})(8 x - 2)=\left(8 x - 2\right) \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} + \left(12 x + 9\right) \left(4 x^{2} - 2 x + 2\right) \log{\left(x \right)} + \frac{\left(4 x^{2} - 2 x + 2\right) \left(6 x^{2} + 9 x - 9\right)}{x}\)
\begin{question}Find the derivative of $y = (4 x^{2} - 2 x + 2)(\log{\left(x \right)})(6 x^{2} + 9 x - 9)$.
\soln{9cm}{Identifying $f=4 x^{2} - 2 x + 2$ and $g=\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)}$ and using the product rule with $f=4 x^{2} - 2 x + 2 \implies f'=8 x - 2$. This leaves g as $g = \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$ and $g=6 x^{2} + 9 x - 9 \implies g'=12 x + 9$. Popping up a level gives $g'=(6 x^{2} + 9 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(12 x + 9)$Popping up again (Back to the original problem) gives $f'=(4 x^{2} - 2 x + 2)(\left(12 x + 9\right) \log{\left(x \right)} + \frac{6 x^{2} + 9 x - 9}{x})+(\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)})(8 x - 2)=\left(8 x - 2\right) \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} + \left(12 x + 9\right) \left(4 x^{2} - 2 x + 2\right) \log{\left(x \right)} + \frac{\left(4 x^{2} - 2 x + 2\right) \left(6 x^{2} + 9 x - 9\right)}{x}$}
\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 = (4 x^{2} - 2 x + 2)(\log{\left(x \right)})(6 x^{2} + 9 x - 9) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%284%20x%5E%7B2%7D%20-%202%20x%20%2B%202%29%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%29%20" alt="LaTeX: \displaystyle y = (4 x^{2} - 2 x + 2)(\log{\left(x \right)})(6 x^{2} + 9 x - 9) " data-equation-content=" \displaystyle y = (4 x^{2} - 2 x + 2)(\log{\left(x \right)})(6 x^{2} + 9 x - 9) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=4 x^{2} - 2 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20x%5E%7B2%7D%20-%202%20x%20%2B%202%20" alt="LaTeX: \displaystyle f=4 x^{2} - 2 x + 2 " data-equation-content=" \displaystyle f=4 x^{2} - 2 x + 2 " /> and <img class="equation_image" title=" \displaystyle g=\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g=\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=4 x^{2} - 2 x + 2 \implies f'=8 x - 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20x%5E%7B2%7D%20-%202%20x%20%2B%202%20%5Cimplies%20f%27%3D8%20x%20-%202%20" alt="LaTeX: \displaystyle f=4 x^{2} - 2 x + 2 \implies f'=8 x - 2 " data-equation-content=" \displaystyle f=4 x^{2} - 2 x + 2 \implies f'=8 x - 2 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " data-equation-content=" \displaystyle g = \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " data-equation-content=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " /> and <img class="equation_image" title=" \displaystyle g=6 x^{2} + 9 x - 9 \implies g'=12 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20g%3D6%20x%5E%7B2%7D%20%2B%209%20x%20-%209%20%5Cimplies%20g%27%3D12%20x%20%2B%209%20" alt="LaTeX: \displaystyle g=6 x^{2} + 9 x - 9 \implies g'=12 x + 9 " data-equation-content=" \displaystyle g=6 x^{2} + 9 x - 9 \implies g'=12 x + 9 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(6 x^{2} + 9 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(12 x + 9) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%29%28%5Cfrac%7B1%7D%7Bx%7D%29%2B%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%2812%20x%20%2B%209%29%20" alt="LaTeX: \displaystyle g'=(6 x^{2} + 9 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(12 x + 9) " data-equation-content=" \displaystyle g'=(6 x^{2} + 9 x - 9)(\frac{1}{x})+(\log{\left(x \right)})(12 x + 9) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(4 x^{2} - 2 x + 2)(\left(12 x + 9\right) \log{\left(x \right)} + \frac{6 x^{2} + 9 x - 9}{x})+(\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)})(8 x - 2)=\left(8 x - 2\right) \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} + \left(12 x + 9\right) \left(4 x^{2} - 2 x + 2\right) \log{\left(x \right)} + \frac{\left(4 x^{2} - 2 x + 2\right) \left(6 x^{2} + 9 x - 9\right)}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%284%20x%5E%7B2%7D%20-%202%20x%20%2B%202%29%28%5Cleft%2812%20x%20%2B%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B6%20x%5E%7B2%7D%20%2B%209%20x%20-%209%7D%7Bx%7D%29%2B%28%5Cleft%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%288%20x%20-%202%29%3D%5Cleft%288%20x%20-%202%5Cright%29%20%5Cleft%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%20%2B%209%5Cright%29%20%5Cleft%284%20x%5E%7B2%7D%20-%202%20x%20%2B%202%5Cright%29%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%284%20x%5E%7B2%7D%20-%202%20x%20%2B%202%5Cright%29%20%5Cleft%286%20x%5E%7B2%7D%20%2B%209%20x%20-%209%5Cright%29%7D%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(4 x^{2} - 2 x + 2)(\left(12 x + 9\right) \log{\left(x \right)} + \frac{6 x^{2} + 9 x - 9}{x})+(\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)})(8 x - 2)=\left(8 x - 2\right) \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} + \left(12 x + 9\right) \left(4 x^{2} - 2 x + 2\right) \log{\left(x \right)} + \frac{\left(4 x^{2} - 2 x + 2\right) \left(6 x^{2} + 9 x - 9\right)}{x} " data-equation-content=" \displaystyle f'=(4 x^{2} - 2 x + 2)(\left(12 x + 9\right) \log{\left(x \right)} + \frac{6 x^{2} + 9 x - 9}{x})+(\left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)})(8 x - 2)=\left(8 x - 2\right) \left(6 x^{2} + 9 x - 9\right) \log{\left(x \right)} + \left(12 x + 9\right) \left(4 x^{2} - 2 x + 2\right) \log{\left(x \right)} + \frac{\left(4 x^{2} - 2 x + 2\right) \left(6 x^{2} + 9 x - 9\right)}{x} " /> </p> </p>