Find the derivative of LaTeX: \displaystyle y = \frac{x^{7} \left(9 x + 8\right)^{7} e^{- x}}{\left(x + 1\right)^{8} \left(8 x + 8\right)^{8}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{x^{7} \left(9 x + 8\right)^{7} e^{- x}}{\left(x + 1\right)^{8} \left(8 x + 8\right)^{8}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(x \right)} + 7 \ln{\left(9 x + 8 \right)}- x - 8 \ln{\left(x + 1 \right)} - 8 \ln{\left(8 x + 8 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{63}{9 x + 8} - \frac{64}{8 x + 8} - \frac{8}{x + 1} + \frac{7}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{63}{9 x + 8} - \frac{64}{8 x + 8} - \frac{8}{x + 1} + \frac{7}{x}\right)\left(\frac{x^{7} \left(9 x + 8\right)^{7} e^{- x}}{\left(x + 1\right)^{8} \left(8 x + 8\right)^{8}} \right)