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Calculus
Derivatives
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Find the derivative of \(\displaystyle y = \left(6 x^{2} + 2 x + 9\right) e^{- x}\).

Using the quotient rule with \(\displaystyle f = 6 x^{2} + 2 x + 9\), \(\displaystyle f' = 12 x + 2\), \(\displaystyle g = e^{x}\), and \(\displaystyle g'= e^{x}\) gives:
\begin{equation*} \frac{ (e^{x})(12 x + 2) - (6 x^{2} + 2 x + 9)(e^{x})}{(e^{x})^2} = - \left(6 x^{2} - 10 x + 7\right) e^{- x} \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the derivative of $y = \left(6 x^{2} + 2 x + 9\right) e^{- x}$. 
    \soln{9cm}{Using the quotient rule with $f = 6 x^{2} + 2 x + 9$, $f' = 12 x + 2$, $g = e^{x}$, and $g'= e^{x}$ gives:\newline 
 \begin{equation*} \frac{ (e^{x})(12 x + 2) - (6 x^{2} + 2 x + 9)(e^{x})}{(e^{x})^2} = - \left(6 x^{2} - 10 x + 7\right) e^{- x} \end{equation*}}

\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 = \left(6 x^{2} + 2 x + 9\right) e^{- x} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cleft%286%20x%5E%7B2%7D%20%2B%202%20x%20%2B%209%5Cright%29%20e%5E%7B-%20x%7D%20" alt="LaTeX:  \displaystyle y = \left(6 x^{2} + 2 x + 9\right) e^{- x} " data-equation-content=" \displaystyle y = \left(6 x^{2} + 2 x + 9\right) e^{- x} " /> . </p> </p>
HTML for Canvas
<p> <p>Using the quotient rule with  <img class="equation_image" title=" \displaystyle f = 6 x^{2} + 2 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%20%3D%206%20x%5E%7B2%7D%20%2B%202%20x%20%2B%209%20" alt="LaTeX:  \displaystyle f = 6 x^{2} + 2 x + 9 " data-equation-content=" \displaystyle f = 6 x^{2} + 2 x + 9 " /> ,  <img class="equation_image" title=" \displaystyle f' = 12 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%27%20%3D%2012%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f' = 12 x + 2 " data-equation-content=" \displaystyle f' = 12 x + 2 " /> ,  <img class="equation_image" title=" \displaystyle g = e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = e^{x} " data-equation-content=" \displaystyle g = e^{x} " /> , and  <img class="equation_image" title=" \displaystyle g'= e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g'= e^{x} " data-equation-content=" \displaystyle g'= e^{x} " />  gives:<br> 
  <img class="equation_image" title="  \frac{ (e^{x})(12 x + 2) - (6 x^{2} + 2 x + 9)(e^{x})}{(e^{x})^2} = - \left(6 x^{2} - 10 x + 7\right) e^{- x}  " src="/equation_images/%20%20%5Cfrac%7B%20%28e%5E%7Bx%7D%29%2812%20x%20%2B%202%29%20-%20%286%20x%5E%7B2%7D%20%2B%202%20x%20%2B%209%29%28e%5E%7Bx%7D%29%7D%7B%28e%5E%7Bx%7D%29%5E2%7D%20%3D%20-%20%5Cleft%286%20x%5E%7B2%7D%20-%2010%20x%20%2B%207%5Cright%29%20e%5E%7B-%20x%7D%20%20" alt="LaTeX:   \frac{ (e^{x})(12 x + 2) - (6 x^{2} + 2 x + 9)(e^{x})}{(e^{x})^2} = - \left(6 x^{2} - 10 x + 7\right) e^{- x}  " data-equation-content="  \frac{ (e^{x})(12 x + 2) - (6 x^{2} + 2 x + 9)(e^{x})}{(e^{x})^2} = - \left(6 x^{2} - 10 x + 7\right) e^{- x}  " /> </p> </p>