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Calculus
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Find the derivative of \(\displaystyle y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}}\)

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: \begin{equation*}\ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)} \end{equation*} Expanding the right hand side using the product and quotient properties of logarithms gives: \begin{equation*}\ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)} \end{equation*} Taking the derivative on both sides of the equation yields: \begin{equation*}\frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9} \end{equation*} Solving for \(\displaystyle y'\) and substituting out y using the original equation gives \begin{equation*}y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right) \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the derivative of $y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}}$
    \soln{9cm}{Taking the natural logarithm of both sides of the equation and expanding the right hand side gives:
\begin{equation*}\ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)}  \end{equation*}
Expanding the right hand side using the product and quotient properties of logarithms gives:
\begin{equation*}\ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)}  \end{equation*}
Taking the derivative on both sides of the equation yields:
\begin{equation*}\frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}  \end{equation*}
Solving for $y'$ and substituting out y using the original equation gives
\begin{equation*}y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)  \end{equation*}
}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

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HTML for Canvas 
<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B%5Cleft%289%20x%20%2B%206%5Cright%29%5E%7B2%7D%20%5Csqrt%7B%5Cleft%285%20x%20%2B%202%5Cright%29%5E%7B3%7D%7D%20e%5E%7B-%20x%7D%7D%7B%5Cleft%28-%205%20x%20-%209%5Cright%29%5E%7B5%7D%20%5Cleft%28x%20%2B%202%5Cright%29%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} " data-equation-content=" \displaystyle y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} " /> </p> </p>
HTML for Canvas
<p> <p>Taking the natural logarithm of both sides of the equation and expanding the right hand side gives:
 <img class="equation_image" title=" \ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)}   " src="/equation_images/%20%5Cln%28y%29%20%3D%20%5Cln%7B%5Cleft%28%5Cfrac%7B%5Cleft%289%20x%20%2B%206%5Cright%29%5E%7B2%7D%20%5Csqrt%7B%5Cleft%285%20x%20%2B%202%5Cright%29%5E%7B3%7D%7D%20e%5E%7B-%20x%7D%7D%7B%5Cleft%28-%205%20x%20-%209%5Cright%29%5E%7B5%7D%20%5Cleft%28x%20%2B%202%5Cright%29%5E%7B2%7D%7D%20%5Cright%29%7D%20%20%20" alt="LaTeX:  \ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)}   " data-equation-content=" \ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)}   " /> 
Expanding the right hand side using the product and quotient properties of logarithms gives:
 <img class="equation_image" title=" \ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)}   " src="/equation_images/%20%5Cln%28y%29%20%3D%20%5Cfrac%7B3%20%5Cln%7B%5Cleft%285%20x%20%2B%202%20%5Cright%29%7D%7D%7B2%7D%20%2B%202%20%5Cln%7B%5Cleft%289%20x%20%2B%206%20%5Cright%29%7D-%20x%20-%205%20%5Cln%7B%5Cleft%28-%205%20x%20-%209%20%5Cright%29%7D%20-%202%20%5Cln%7B%5Cleft%28x%20%2B%202%20%5Cright%29%7D%20%20%20" alt="LaTeX:  \ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)}   " data-equation-content=" \ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)}   " /> 
Taking the derivative on both sides of the equation yields:
 <img class="equation_image" title=" \frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}   " src="/equation_images/%20%5Cfrac%7By%27%7D%7By%7D%20%3D%20-1%20%2B%20%5Cfrac%7B18%7D%7B9%20x%20%2B%206%7D%20%2B%20%5Cfrac%7B15%7D%7B2%20%5Cleft%285%20x%20%2B%202%5Cright%29%7D%20-%20%5Cfrac%7B2%7D%7Bx%20%2B%202%7D%20%2B%20%5Cfrac%7B25%7D%7B-%205%20x%20-%209%7D%20%20%20" alt="LaTeX:  \frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}   " data-equation-content=" \frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}   " /> 
Solving for  <img class="equation_image" title=" \displaystyle y' " src="/equation_images/%20%5Cdisplaystyle%20y%27%20" alt="LaTeX:  \displaystyle y' " data-equation-content=" \displaystyle y' " />  and substituting out y using the original equation gives
 <img class="equation_image" title=" y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)   " src="/equation_images/%20y%27%20%3D%20%5Cleft%28-1%20%2B%20%5Cfrac%7B18%7D%7B9%20x%20%2B%206%7D%20%2B%20%5Cfrac%7B15%7D%7B2%20%5Cleft%285%20x%20%2B%202%5Cright%29%7D%20-%20%5Cfrac%7B2%7D%7Bx%20%2B%202%7D%20%2B%20%5Cfrac%7B25%7D%7B-%205%20x%20-%209%7D%5Cright%29%5Cleft%28%5Cfrac%7B%5Cleft%289%20x%20%2B%206%5Cright%29%5E%7B2%7D%20%5Csqrt%7B%5Cleft%285%20x%20%2B%202%5Cright%29%5E%7B3%7D%7D%20e%5E%7B-%20x%7D%7D%7B%5Cleft%28-%205%20x%20-%209%5Cright%29%5E%7B5%7D%20%5Cleft%28x%20%2B%202%5Cright%29%5E%7B2%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX:  y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)   " data-equation-content=" y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)   " /> 
</p> </p>