Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(8 x - 9\right)^{8} \left(9 x - 4\right)^{8} \sqrt{\left(4 x + 8\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 + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(8 x - 9\right)^{8} \left(9 x - 4\right)^{8} \sqrt{\left(4 x + 8\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(2 x + 8 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- x - \frac{5 \ln{\left(4 x + 8 \right)}}{2} - 8 \ln{\left(8 x - 9 \right)} - 8 \ln{\left(9 x - 4 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{72}{9 x - 4} - \frac{64}{8 x - 9} - \frac{10}{4 x + 8} + \frac{4}{2 x + 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{72}{9 x - 4} - \frac{64}{8 x - 9} - \frac{10}{4 x + 8} + \frac{4}{2 x + 8}\right)\left(\frac{\left(2 x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(8 x - 9\right)^{8} \left(9 x - 4\right)^{8} \sqrt{\left(4 x + 8\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{5}{\tan{\left(x \right)}} + \frac{4}{2 x + 8}-1 - \frac{72}{9 x - 4} - \frac{64}{8 x - 9} - \frac{10}{4 x + 8}\right)\left(\frac{\left(2 x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(8 x - 9\right)^{8} \left(9 x - 4\right)^{8} \sqrt{\left(4 x + 8\right)^{5}}} \right)