Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 9\right)^{8} \left(3 x + 9\right)^{2} e^{- x}}{\left(- 4 x - 5\right)^{7} \sqrt{\left(8 x + 9\right)^{7}} \sin^{2}{\left(x \right)}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x + 9\right)^{8} \left(3 x + 9\right)^{2} e^{- x}}{\left(- 4 x - 5\right)^{7} \sqrt{\left(8 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(x + 9 \right)} + 2 \ln{\left(3 x + 9 \right)}- x - 7 \ln{\left(- 4 x - 5 \right)} - \frac{7 \ln{\left(8 x + 9 \right)}}{2} - 2 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 9} + \frac{6}{3 x + 9} + \frac{8}{x + 9} + \frac{28}{- 4 x - 5} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 9} + \frac{6}{3 x + 9} + \frac{8}{x + 9} + \frac{28}{- 4 x - 5}\right)\left(\frac{\left(x + 9\right)^{8} \left(3 x + 9\right)^{2} e^{- x}}{\left(- 4 x - 5\right)^{7} \sqrt{\left(8 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{6}{3 x + 9} + \frac{8}{x + 9}-1 - \frac{2}{\tan{\left(x \right)}} - \frac{28}{8 x + 9} + \frac{28}{- 4 x - 5}\right)\left(\frac{\left(x + 9\right)^{8} \left(3 x + 9\right)^{2} e^{- x}}{\left(- 4 x - 5\right)^{7} \sqrt{\left(8 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right)