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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(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)} \end{equation*} Expanding the right hand side using the product and quotient properties of logarithms gives: \begin{equation*}\ln(y) = x + 5 \ln{\left(5 x + 6 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(5 x - 2 \right)} - 2 \ln{\left(7 \right)} \end{equation*} Taking the derivative on both sides of the equation yields: \begin{equation*}\frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x} \end{equation*} Solving for \(\displaystyle y'\) and substituting out y using the original equation gives \begin{equation*}y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right) \end{equation*} Using some Trigonometric identities to simplify gives \begin{equation*}y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{25}{5 x + 6}- \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right) \end{equation*}

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\begin{question}Find the derivative of $y = \frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}}$
    \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(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)}  \end{equation*}
Expanding the right hand side using the product and quotient properties of logarithms gives:
\begin{equation*}\ln(y) = x + 5 \ln{\left(5 x + 6 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(5 x - 2 \right)} - 2 \ln{\left(7 \right)}  \end{equation*}
Taking the derivative on both sides of the equation yields:
\begin{equation*}\frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}  \end{equation*}
Solving for $y'$ and substituting out y using the original equation gives
\begin{equation*}y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)  \end{equation*}
Using some Trigonometric identities to simplify gives
\begin{equation*}y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{25}{5 x + 6}- \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)  \end{equation*}
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\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.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = \frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B%5Cleft%285%20x%20%2B%206%5Cright%29%5E%7B5%7D%20e%5E%7Bx%7D%20%5Ccos%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B49%20x%5E%7B2%7D%20%5Cleft%285%20x%20-%202%5Cright%29%5E%7B8%7D%7D%20" alt="LaTeX:  \displaystyle y = \frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} " data-equation-content=" \displaystyle y = \frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} " /> </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(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)}   " src="/equation_images/%20%5Cln%28y%29%20%3D%20%5Cln%7B%5Cleft%28%5Cfrac%7B%5Cleft%285%20x%20%2B%206%5Cright%29%5E%7B5%7D%20e%5E%7Bx%7D%20%5Ccos%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B49%20x%5E%7B2%7D%20%5Cleft%285%20x%20-%202%5Cright%29%5E%7B8%7D%7D%20%5Cright%29%7D%20%20%20" alt="LaTeX:  \ln(y) = \ln{\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)}   " data-equation-content=" \ln(y) = \ln{\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)}   " /> 
Expanding the right hand side using the product and quotient properties of logarithms gives:
 <img class="equation_image" title=" \ln(y) = x + 5 \ln{\left(5 x + 6 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(5 x - 2 \right)} - 2 \ln{\left(7 \right)}   " src="/equation_images/%20%5Cln%28y%29%20%3D%20x%20%2B%205%20%5Cln%7B%5Cleft%285%20x%20%2B%206%20%5Cright%29%7D%20%2B%207%20%5Cln%7B%5Cleft%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D-%202%20%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20-%208%20%5Cln%7B%5Cleft%285%20x%20-%202%20%5Cright%29%7D%20-%202%20%5Cln%7B%5Cleft%287%20%5Cright%29%7D%20%20%20" alt="LaTeX:  \ln(y) = x + 5 \ln{\left(5 x + 6 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(5 x - 2 \right)} - 2 \ln{\left(7 \right)}   " data-equation-content=" \ln(y) = x + 5 \ln{\left(5 x + 6 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(5 x - 2 \right)} - 2 \ln{\left(7 \right)}   " /> 
Taking the derivative on both sides of the equation yields:
 <img class="equation_image" title=" \frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}   " src="/equation_images/%20%5Cfrac%7By%27%7D%7By%7D%20%3D%20-%20%5Cfrac%7B7%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%2B%201%20%2B%20%5Cfrac%7B25%7D%7B5%20x%20%2B%206%7D%20-%20%5Cfrac%7B40%7D%7B5%20x%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bx%7D%20%20%20" alt="LaTeX:  \frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}   " data-equation-content=" \frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}   " /> 
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(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " src="/equation_images/%20y%27%20%3D%20%5Cleft%28-%20%5Cfrac%7B7%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%2B%201%20%2B%20%5Cfrac%7B25%7D%7B5%20x%20%2B%206%7D%20-%20%5Cfrac%7B40%7D%7B5%20x%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bx%7D%5Cright%29%5Cleft%28%5Cfrac%7B%5Cleft%285%20x%20%2B%206%5Cright%29%5E%7B5%7D%20e%5E%7Bx%7D%20%5Ccos%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B49%20x%5E%7B2%7D%20%5Cleft%285%20x%20-%202%5Cright%29%5E%7B8%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX:  y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " data-equation-content=" y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{25}{5 x + 6} - \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " /> 
Using some Trigonometric identities to simplify gives
 <img class="equation_image" title=" y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{25}{5 x + 6}- \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " src="/equation_images/%20y%27%20%3D%20%5Cleft%28-%207%20%5Ctan%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%201%20%2B%20%5Cfrac%7B25%7D%7B5%20x%20%2B%206%7D-%20%5Cfrac%7B40%7D%7B5%20x%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bx%7D%5Cright%29%5Cleft%28%5Cfrac%7B%5Cleft%285%20x%20%2B%206%5Cright%29%5E%7B5%7D%20e%5E%7Bx%7D%20%5Ccos%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B49%20x%5E%7B2%7D%20%5Cleft%285%20x%20-%202%5Cright%29%5E%7B8%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX:  y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{25}{5 x + 6}- \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " data-equation-content=" y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{25}{5 x + 6}- \frac{40}{5 x - 2} - \frac{2}{x}\right)\left(\frac{\left(5 x + 6\right)^{5} e^{x} \cos^{7}{\left(x \right)}}{49 x^{2} \left(5 x - 2\right)^{8}} \right)   " /> 
</p> </p>