Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 2\right)^{2} \sqrt{\left(3 x + 9\right)^{3}} e^{- x}}{\left(8 x - 6\right)^{6} \sin^{5}{\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 - 2\right)^{2} \sqrt{\left(3 x + 9\right)^{3}} e^{- x}}{\left(8 x - 6\right)^{6} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x - 2 \right)} + \frac{3 \ln{\left(3 x + 9 \right)}}{2}- x - 2 \ln{\left(x - 2 \right)} - 6 \ln{\left(8 x - 6 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{48}{8 x - 6} + \frac{9}{2 \left(3 x + 9\right)} + \frac{2}{x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{48}{8 x - 6} + \frac{9}{2 \left(3 x + 9\right)} + \frac{2}{x - 2}\right)\left(\frac{\left(x - 2\right)^{2} \sqrt{\left(3 x + 9\right)^{3}} e^{- x}}{\left(8 x - 6\right)^{6} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{9}{2 \left(3 x + 9\right)} + \frac{4}{x - 2}-1 - \frac{5}{\tan{\left(x \right)}} - \frac{48}{8 x - 6} - \frac{2}{x - 2}\right)\left(\frac{\left(x - 2\right)^{2} \sqrt{\left(3 x + 9\right)^{3}} e^{- x}}{\left(8 x - 6\right)^{6} \sin^{5}{\left(x \right)}} \right)