Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 x - 8\right)^{3} e^{- x} \sin^{8}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{6} \left(9 x + 1\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(4 x - 8\right)^{3} e^{- x} \sin^{8}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{6} \left(9 x + 1\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(4 x - 8 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(- 4 x - 8 \right)} - 2 \ln{\left(9 x + 1 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{9 x + 1} + \frac{12}{4 x - 8} + \frac{24}{- 4 x - 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{9 x + 1} + \frac{12}{4 x - 8} + \frac{24}{- 4 x - 8}\right)\left(\frac{\left(4 x - 8\right)^{3} e^{- x} \sin^{8}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{6} \left(9 x + 1\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{8}{\tan{\left(x \right)}} + \frac{12}{4 x - 8}-1 - \frac{18}{9 x + 1} + \frac{24}{- 4 x - 8}\right)\left(\frac{\left(4 x - 8\right)^{3} e^{- x} \sin^{8}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{6} \left(9 x + 1\right)^{2}} \right)