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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (9 x - 3)(e^{x})(1 - 9 x)\).
Identifying \(\displaystyle f=9 x - 3\) and \(\displaystyle g=\left(1 - 9 x\right) e^{x}\) and using the product rule with \(\displaystyle f=9 x - 3 \implies f'=9\). This leaves g as \(\displaystyle g = \left(1 - 9 x\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=e^{x} \implies f'=e^{x}\) and \(\displaystyle g=1 - 9 x \implies g'=-9\). Popping up a level gives \(\displaystyle g'=(1 - 9 x)(e^{x})+(e^{x})(-9)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(9 x - 3)(\left(1 - 9 x\right) e^{x} - 9 e^{x})+(\left(1 - 9 x\right) e^{x})(9)=\left(1 - 9 x\right) \left(9 x - 3\right) e^{x} + \left(9 - 81 x\right) e^{x} + \left(27 - 81 x\right) e^{x}\)
\begin{question}Find the derivative of $y = (9 x - 3)(e^{x})(1 - 9 x)$.
\soln{9cm}{Identifying $f=9 x - 3$ and $g=\left(1 - 9 x\right) e^{x}$ and using the product rule with $f=9 x - 3 \implies f'=9$. This leaves g as $g = \left(1 - 9 x\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=e^{x} \implies f'=e^{x}$ and $g=1 - 9 x \implies g'=-9$. Popping up a level gives $g'=(1 - 9 x)(e^{x})+(e^{x})(-9)$Popping up again (Back to the original problem) gives $f'=(9 x - 3)(\left(1 - 9 x\right) e^{x} - 9 e^{x})+(\left(1 - 9 x\right) e^{x})(9)=\left(1 - 9 x\right) \left(9 x - 3\right) e^{x} + \left(9 - 81 x\right) e^{x} + \left(27 - 81 x\right) e^{x}$}
\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 = (9 x - 3)(e^{x})(1 - 9 x) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%289%20x%20-%203%29%28e%5E%7Bx%7D%29%281%20-%209%20x%29%20" alt="LaTeX: \displaystyle y = (9 x - 3)(e^{x})(1 - 9 x) " data-equation-content=" \displaystyle y = (9 x - 3)(e^{x})(1 - 9 x) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=9 x - 3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D9%20x%20-%203%20" alt="LaTeX: \displaystyle f=9 x - 3 " data-equation-content=" \displaystyle f=9 x - 3 " /> and <img class="equation_image" title=" \displaystyle g=\left(1 - 9 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%281%20-%209%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g=\left(1 - 9 x\right) e^{x} " data-equation-content=" \displaystyle g=\left(1 - 9 x\right) e^{x} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=9 x - 3 \implies f'=9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D9%20x%20-%203%20%5Cimplies%20f%27%3D9%20" alt="LaTeX: \displaystyle f=9 x - 3 \implies f'=9 " data-equation-content=" \displaystyle f=9 x - 3 \implies f'=9 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(1 - 9 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%281%20-%209%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g = \left(1 - 9 x\right) e^{x} " data-equation-content=" \displaystyle g = \left(1 - 9 x\right) e^{x} " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=e^{x} \implies f'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20%5Cimplies%20f%27%3De%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f=e^{x} \implies f'=e^{x} " data-equation-content=" \displaystyle f=e^{x} \implies f'=e^{x} " /> and <img class="equation_image" title=" \displaystyle g=1 - 9 x \implies g'=-9 " src="/equation_images/%20%5Cdisplaystyle%20g%3D1%20-%209%20x%20%5Cimplies%20g%27%3D-9%20" alt="LaTeX: \displaystyle g=1 - 9 x \implies g'=-9 " data-equation-content=" \displaystyle g=1 - 9 x \implies g'=-9 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(1 - 9 x)(e^{x})+(e^{x})(-9) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%281%20-%209%20x%29%28e%5E%7Bx%7D%29%2B%28e%5E%7Bx%7D%29%28-9%29%20" alt="LaTeX: \displaystyle g'=(1 - 9 x)(e^{x})+(e^{x})(-9) " data-equation-content=" \displaystyle g'=(1 - 9 x)(e^{x})+(e^{x})(-9) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(9 x - 3)(\left(1 - 9 x\right) e^{x} - 9 e^{x})+(\left(1 - 9 x\right) e^{x})(9)=\left(1 - 9 x\right) \left(9 x - 3\right) e^{x} + \left(9 - 81 x\right) e^{x} + \left(27 - 81 x\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%289%20x%20-%203%29%28%5Cleft%281%20-%209%20x%5Cright%29%20e%5E%7Bx%7D%20-%209%20e%5E%7Bx%7D%29%2B%28%5Cleft%281%20-%209%20x%5Cright%29%20e%5E%7Bx%7D%29%289%29%3D%5Cleft%281%20-%209%20x%5Cright%29%20%5Cleft%289%20x%20-%203%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%289%20-%2081%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2827%20-%2081%20x%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle f'=(9 x - 3)(\left(1 - 9 x\right) e^{x} - 9 e^{x})+(\left(1 - 9 x\right) e^{x})(9)=\left(1 - 9 x\right) \left(9 x - 3\right) e^{x} + \left(9 - 81 x\right) e^{x} + \left(27 - 81 x\right) e^{x} " data-equation-content=" \displaystyle f'=(9 x - 3)(\left(1 - 9 x\right) e^{x} - 9 e^{x})+(\left(1 - 9 x\right) e^{x})(9)=\left(1 - 9 x\right) \left(9 x - 3\right) e^{x} + \left(9 - 81 x\right) e^{x} + \left(27 - 81 x\right) e^{x} " /> </p> </p>