Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 2 \ln{\left(9 x + 6 \right)}- x - 5 \ln{\left(- 5 x - 9 \right)} - 2 \ln{\left(x + 2 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{18}{9 x + 6} + \frac{15}{2 \left(5 x + 2\right)} - \frac{2}{x + 2} + \frac{25}{- 5 x - 9}\right)\left(\frac{\left(9 x + 6\right)^{2} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(- 5 x - 9\right)^{5} \left(x + 2\right)^{2}} \right)