Find the derivative of $LaTeX: \displaystyle y = \frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 7 \ln{\left(4 x + 2 \right)}- 5 \ln{\left(7 x + 1 \right)}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 - \frac{35}{7 x + 1} + \frac{28}{4 x + 2}\right)\left(\frac{\left(4 x + 2\right)^{7} e^{x}}{\left(7 x + 1\right)^{5}} \right)$