Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 4 x - 3\right)^{7} \left(6 x + 3\right)^{4} e^{x}}{\left(x + 4\right)^{4} \left(4 x + 5\right)^{4}}

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 - 3\right)^{7} \left(6 x + 3\right)^{4} e^{x}}{\left(x + 4\right)^{4} \left(4 x + 5\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(- 4 x - 3 \right)} + 4 \ln{\left(6 x + 3 \right)}- 4 \ln{\left(x + 4 \right)} - 4 \ln{\left(4 x + 5 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{24}{6 x + 3} - \frac{16}{4 x + 5} - \frac{4}{x + 4} - \frac{28}{- 4 x - 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{24}{6 x + 3} - \frac{16}{4 x + 5} - \frac{4}{x + 4} - \frac{28}{- 4 x - 3}\right)\left(\frac{\left(- 4 x - 3\right)^{7} \left(6 x + 3\right)^{4} e^{x}}{\left(x + 4\right)^{4} \left(4 x + 5\right)^{4}} \right)