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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}}\)
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(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)} \end{equation*} Expanding the right hand side using the product and quotient properties of logarithms gives: \begin{equation*}\ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)} \end{equation*} Taking the derivative on both sides of the equation yields: \begin{equation*}\frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2} \end{equation*} Solving for \(\displaystyle y'\) and substituting out y using the original equation gives \begin{equation*}y' = \left(1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right) \end{equation*}
\begin{question}Find the derivative of $y = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}}$
\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(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)} \end{equation*}
Expanding the right hand side using the product and quotient properties of logarithms gives:
\begin{equation*}\ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)} \end{equation*}
Taking the derivative on both sides of the equation yields:
\begin{equation*}\frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2} \end{equation*}
Solving for $y'$ and substituting out y using the original equation gives
\begin{equation*}y' = \left(1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right) \end{equation*}
}
\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 = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B%5Cleft%284%20x%20%2B%202%5Cright%29%5E%7B7%7D%20e%5E%7Bx%7D%7D%7B%5Cleft%287%20x%20%2B%201%5Cright%29%5E%7B5%7D%7D%20" alt="LaTeX: \displaystyle y = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} " data-equation-content=" \displaystyle y = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} " /> </p> </p><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(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)} " src="/equation_images/%20%5Cln%28y%29%20%3D%20%5Cln%7B%5Cleft%28%5Cfrac%7B%5Cleft%284%20x%20%2B%202%5Cright%29%5E%7B7%7D%20e%5E%7Bx%7D%7D%7B%5Cleft%287%20x%20%2B%201%5Cright%29%5E%7B5%7D%7D%20%5Cright%29%7D%20%20%20" alt="LaTeX: \ln(y) = \ln{\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)} " data-equation-content=" \ln(y) = \ln{\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)} " />
Expanding the right hand side using the product and quotient properties of logarithms gives:
<img class="equation_image" title=" \ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)} " src="/equation_images/%20%5Cln%28y%29%20%3D%20x%20%2B%207%20%5Cln%7B%5Cleft%284%20x%20%2B%202%20%5Cright%29%7D-%205%20%5Cln%7B%5Cleft%287%20x%20%2B%201%20%5Cright%29%7D%20%20%20" alt="LaTeX: \ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)} " data-equation-content=" \ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)} " />
Taking the derivative on both sides of the equation yields:
<img class="equation_image" title=" \frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2} " src="/equation_images/%20%5Cfrac%7By%27%7D%7By%7D%20%3D%201%20-%20%5Cfrac%7B35%7D%7B7%20x%20%2B%201%7D%20%2B%20%5Cfrac%7B28%7D%7B4%20x%20%2B%202%7D%20%20%20" alt="LaTeX: \frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2} " data-equation-content=" \frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2} " />
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{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right) " src="/equation_images/%20y%27%20%3D%20%5Cleft%281%20-%20%5Cfrac%7B35%7D%7B7%20x%20%2B%201%7D%20%2B%20%5Cfrac%7B28%7D%7B4%20x%20%2B%202%7D%5Cright%29%5Cleft%28%5Cfrac%7B%5Cleft%284%20x%20%2B%202%5Cright%29%5E%7B7%7D%20e%5E%7Bx%7D%7D%7B%5Cleft%287%20x%20%2B%201%5Cright%29%5E%7B5%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX: y' = \left(1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right) " data-equation-content=" y' = \left(1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right) " />
</p> </p>