Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 x - 1\right)^{5} \sqrt{\left(3 x + 8\right)^{3}} e^{- x}}{\left(x - 4\right)^{4} \left(x - 2\right)^{2} \left(5 x + 2\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(2 x - 1\right)^{5} \sqrt{\left(3 x + 8\right)^{3}} e^{- x}}{\left(x - 4\right)^{4} \left(x - 2\right)^{2} \left(5 x + 2\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(2 x - 1 \right)} + \frac{3 \ln{\left(3 x + 8 \right)}}{2}- x - 4 \ln{\left(x - 4 \right)} - 2 \ln{\left(x - 2 \right)} - 5 \ln{\left(5 x + 2 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{25}{5 x + 2} + \frac{9}{2 \left(3 x + 8\right)} + \frac{10}{2 x - 1} - \frac{2}{x - 2} - \frac{4}{x - 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{25}{5 x + 2} + \frac{9}{2 \left(3 x + 8\right)} + \frac{10}{2 x - 1} - \frac{2}{x - 2} - \frac{4}{x - 4}\right)\left(\frac{\left(2 x - 1\right)^{5} \sqrt{\left(3 x + 8\right)^{3}} e^{- x}}{\left(x - 4\right)^{4} \left(x - 2\right)^{2} \left(5 x + 2\right)^{5}} \right)