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Calculus
Derivatives
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Find the derivative of \(\displaystyle y = (e^{x})(5 x + 4)(7 - 6 x)\).


Identifying \(\displaystyle f=e^{x}\) and \(\displaystyle g=\left(7 - 6 x\right) \left(5 x + 4\right)\) and using the product rule with \(\displaystyle f=e^{x} \implies f'=e^{x}\). This leaves g as \(\displaystyle g = \left(7 - 6 x\right) \left(5 x + 4\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=5 x + 4 \implies f'=5\) and \(\displaystyle g=7 - 6 x \implies g'=-6\). Popping up a level gives \(\displaystyle g'=(7 - 6 x)(5)+(5 x + 4)(-6)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(e^{x})(11 - 60 x)+(\left(7 - 6 x\right) \left(5 x + 4\right))(e^{x})=\left(7 - 6 x\right) \left(5 x + 4\right) e^{x} + \left(35 - 30 x\right) e^{x} + \left(- 30 x - 24\right) e^{x}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (e^{x})(5 x + 4)(7 - 6 x)$.
    \soln{9cm}{Identifying $f=e^{x}$ and $g=\left(7 - 6 x\right) \left(5 x + 4\right)$ and using the product rule with $f=e^{x} \implies f'=e^{x}$. This leaves g as $g = \left(7 - 6 x\right) \left(5 x + 4\right)$ which also requires the product rule. Pushing down in the new product rule $f=5 x + 4 \implies f'=5$ and $g=7 - 6 x \implies g'=-6$. Popping up a level gives $g'=(7 - 6 x)(5)+(5 x + 4)(-6)$Popping up again (Back to the original problem) gives $f'=(e^{x})(11 - 60 x)+(\left(7 - 6 x\right) \left(5 x + 4\right))(e^{x})=\left(7 - 6 x\right) \left(5 x + 4\right) e^{x} + \left(35 - 30 x\right) e^{x} + \left(- 30 x - 24\right) e^{x}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas
<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (e^{x})(5 x + 4)(7 - 6 x) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28e%5E%7Bx%7D%29%285%20x%20%2B%204%29%287%20-%206%20x%29%20" alt="LaTeX:  \displaystyle y = (e^{x})(5 x + 4)(7 - 6 x) " data-equation-content=" \displaystyle y = (e^{x})(5 x + 4)(7 - 6 x) " /> .</p> </p>
HTML for Canvas
<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f=e^{x} " data-equation-content=" \displaystyle f=e^{x} " />  and  <img class="equation_image" title=" \displaystyle g=\left(7 - 6 x\right) \left(5 x + 4\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%287%20-%206%20x%5Cright%29%20%5Cleft%285%20x%20%2B%204%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(7 - 6 x\right) \left(5 x + 4\right) " data-equation-content=" \displaystyle g=\left(7 - 6 x\right) \left(5 x + 4\right) " />  and using the product rule with  <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} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(7 - 6 x\right) \left(5 x + 4\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%287%20-%206%20x%5Cright%29%20%5Cleft%285%20x%20%2B%204%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(7 - 6 x\right) \left(5 x + 4\right) " data-equation-content=" \displaystyle g = \left(7 - 6 x\right) \left(5 x + 4\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=5 x + 4 \implies f'=5 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20x%20%2B%204%20%5Cimplies%20f%27%3D5%20" alt="LaTeX:  \displaystyle f=5 x + 4 \implies f'=5 " data-equation-content=" \displaystyle f=5 x + 4 \implies f'=5 " />  and  <img class="equation_image" title=" \displaystyle g=7 - 6 x \implies g'=-6 " src="/equation_images/%20%5Cdisplaystyle%20g%3D7%20-%206%20x%20%5Cimplies%20g%27%3D-6%20" alt="LaTeX:  \displaystyle g=7 - 6 x \implies g'=-6 " data-equation-content=" \displaystyle g=7 - 6 x \implies g'=-6 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(7 - 6 x)(5)+(5 x + 4)(-6) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%287%20-%206%20x%29%285%29%2B%285%20x%20%2B%204%29%28-6%29%20" alt="LaTeX:  \displaystyle g'=(7 - 6 x)(5)+(5 x + 4)(-6) " data-equation-content=" \displaystyle g'=(7 - 6 x)(5)+(5 x + 4)(-6) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(e^{x})(11 - 60 x)+(\left(7 - 6 x\right) \left(5 x + 4\right))(e^{x})=\left(7 - 6 x\right) \left(5 x + 4\right) e^{x} + \left(35 - 30 x\right) e^{x} + \left(- 30 x - 24\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28e%5E%7Bx%7D%29%2811%20-%2060%20x%29%2B%28%5Cleft%287%20-%206%20x%5Cright%29%20%5Cleft%285%20x%20%2B%204%5Cright%29%29%28e%5E%7Bx%7D%29%3D%5Cleft%287%20-%206%20x%5Cright%29%20%5Cleft%285%20x%20%2B%204%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%2835%20-%2030%20x%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28-%2030%20x%20-%2024%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(e^{x})(11 - 60 x)+(\left(7 - 6 x\right) \left(5 x + 4\right))(e^{x})=\left(7 - 6 x\right) \left(5 x + 4\right) e^{x} + \left(35 - 30 x\right) e^{x} + \left(- 30 x - 24\right) e^{x} " data-equation-content=" \displaystyle f'=(e^{x})(11 - 60 x)+(\left(7 - 6 x\right) \left(5 x + 4\right))(e^{x})=\left(7 - 6 x\right) \left(5 x + 4\right) e^{x} + \left(35 - 30 x\right) e^{x} + \left(- 30 x - 24\right) e^{x} " /> </p> </p>