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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = (\sin{\left(x \right)})(- 3 x - 2)(e^{x})\).
Identifying \(\displaystyle f=\sin{\left(x \right)}\) and \(\displaystyle g=\left(- 3 x - 2\right) e^{x}\) and using the product rule with \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(- 3 x - 2\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 3 x - 2 \implies f'=-3\) and \(\displaystyle g=e^{x} \implies g'=e^{x}\). Popping up a level gives \(\displaystyle g'=(e^{x})(-3)+(- 3 x - 2)(e^{x})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\sin{\left(x \right)})(\left(- 3 x - 2\right) e^{x} - 3 e^{x})+(\left(- 3 x - 2\right) e^{x})(\cos{\left(x \right)})=\left(- 3 x - 2\right) e^{x} \sin{\left(x \right)} + \left(- 3 x - 2\right) e^{x} \cos{\left(x \right)} - 3 e^{x} \sin{\left(x \right)}\)
\begin{question}Find the derivative of $y = (\sin{\left(x \right)})(- 3 x - 2)(e^{x})$.
\soln{9cm}{Identifying $f=\sin{\left(x \right)}$ and $g=\left(- 3 x - 2\right) e^{x}$ and using the product rule with $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$. This leaves g as $g = \left(- 3 x - 2\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=- 3 x - 2 \implies f'=-3$ and $g=e^{x} \implies g'=e^{x}$. Popping up a level gives $g'=(e^{x})(-3)+(- 3 x - 2)(e^{x})$Popping up again (Back to the original problem) gives $f'=(\sin{\left(x \right)})(\left(- 3 x - 2\right) e^{x} - 3 e^{x})+(\left(- 3 x - 2\right) e^{x})(\cos{\left(x \right)})=\left(- 3 x - 2\right) e^{x} \sin{\left(x \right)} + \left(- 3 x - 2\right) e^{x} \cos{\left(x \right)} - 3 e^{x} \sin{\left(x \right)}$}
\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)})(- 3 x - 2)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%203%20x%20-%202%29%28e%5E%7Bx%7D%29%20" alt="LaTeX: \displaystyle y = (\sin{\left(x \right)})(- 3 x - 2)(e^{x}) " data-equation-content=" \displaystyle y = (\sin{\left(x \right)})(- 3 x - 2)(e^{x}) " /> .</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 - 2\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g=\left(- 3 x - 2\right) e^{x} " data-equation-content=" \displaystyle g=\left(- 3 x - 2\right) e^{x} " /> 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 - 2\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g = \left(- 3 x - 2\right) e^{x} " data-equation-content=" \displaystyle g = \left(- 3 x - 2\right) e^{x} " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=- 3 x - 2 \implies f'=-3 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%203%20x%20-%202%20%5Cimplies%20f%27%3D-3%20" alt="LaTeX: \displaystyle f=- 3 x - 2 \implies f'=-3 " data-equation-content=" \displaystyle f=- 3 x - 2 \implies f'=-3 " /> and <img class="equation_image" title=" \displaystyle g=e^{x} \implies g'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3De%5E%7Bx%7D%20%5Cimplies%20g%27%3De%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g=e^{x} \implies g'=e^{x} " data-equation-content=" \displaystyle g=e^{x} \implies g'=e^{x} " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(e^{x})(-3)+(- 3 x - 2)(e^{x}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28e%5E%7Bx%7D%29%28-3%29%2B%28-%203%20x%20-%202%29%28e%5E%7Bx%7D%29%20" alt="LaTeX: \displaystyle g'=(e^{x})(-3)+(- 3 x - 2)(e^{x}) " data-equation-content=" \displaystyle g'=(e^{x})(-3)+(- 3 x - 2)(e^{x}) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(\sin{\left(x \right)})(\left(- 3 x - 2\right) e^{x} - 3 e^{x})+(\left(- 3 x - 2\right) e^{x})(\cos{\left(x \right)})=\left(- 3 x - 2\right) e^{x} \sin{\left(x \right)} + \left(- 3 x - 2\right) e^{x} \cos{\left(x \right)} - 3 e^{x} \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20-%203%20e%5E%7Bx%7D%29%2B%28%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%203%20x%20-%202%5Cright%29%20e%5E%7Bx%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%203%20e%5E%7Bx%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(\sin{\left(x \right)})(\left(- 3 x - 2\right) e^{x} - 3 e^{x})+(\left(- 3 x - 2\right) e^{x})(\cos{\left(x \right)})=\left(- 3 x - 2\right) e^{x} \sin{\left(x \right)} + \left(- 3 x - 2\right) e^{x} \cos{\left(x \right)} - 3 e^{x} \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(\sin{\left(x \right)})(\left(- 3 x - 2\right) e^{x} - 3 e^{x})+(\left(- 3 x - 2\right) e^{x})(\cos{\left(x \right)})=\left(- 3 x - 2\right) e^{x} \sin{\left(x \right)} + \left(- 3 x - 2\right) e^{x} \cos{\left(x \right)} - 3 e^{x} \sin{\left(x \right)} " /> </p> </p>