Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 - 2 x\right)^{2} \left(9 x - 8\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 4\right)^{4} \left(6 x - 6\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(8 - 2 x\right)^{2} \left(9 x - 8\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 4\right)^{4} \left(6 x - 6\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(8 - 2 x \right)} + 3 \ln{\left(9 x - 8 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- x - 4 \ln{\left(x + 4 \right)} - 2 \ln{\left(6 x - 6 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{9 x - 8} - \frac{12}{6 x - 6} - \frac{4}{x + 4} - \frac{4}{8 - 2 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{9 x - 8} - \frac{12}{6 x - 6} - \frac{4}{x + 4} - \frac{4}{8 - 2 x}\right)\left(\frac{\left(8 - 2 x\right)^{2} \left(9 x - 8\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 4\right)^{4} \left(6 x - 6\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{27}{9 x - 8} - \frac{4}{8 - 2 x}-1 - \frac{12}{6 x - 6} - \frac{4}{x + 4}\right)\left(\frac{\left(8 - 2 x\right)^{2} \left(9 x - 8\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 4\right)^{4} \left(6 x - 6\right)^{2}} \right)