Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 - 8 x\right)^{6} \left(6 x - 6\right)^{3} e^{x} \sin^{8}{\left(x \right)}}{\left(x + 9\right)^{8} \left(9 x + 6\right)^{6}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(6 - 8 x\right)^{6} \left(6 x - 6\right)^{3} e^{x} \sin^{8}{\left(x \right)}}{\left(x + 9\right)^{8} \left(9 x + 6\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(6 - 8 x \right)} + 3 \ln{\left(6 x - 6 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(x + 9 \right)} - 6 \ln{\left(9 x + 6 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{54}{9 x + 6} + \frac{18}{6 x - 6} - \frac{8}{x + 9} - \frac{48}{6 - 8 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{54}{9 x + 6} + \frac{18}{6 x - 6} - \frac{8}{x + 9} - \frac{48}{6 - 8 x}\right)\left(\frac{\left(6 - 8 x\right)^{6} \left(6 x - 6\right)^{3} e^{x} \sin^{8}{\left(x \right)}}{\left(x + 9\right)^{8} \left(9 x + 6\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{8}{\tan{\left(x \right)}} + \frac{18}{6 x - 6} - \frac{48}{6 - 8 x}- \frac{54}{9 x + 6} - \frac{8}{x + 9}\right)\left(\frac{\left(6 - 8 x\right)^{6} \left(6 x - 6\right)^{3} e^{x} \sin^{8}{\left(x \right)}}{\left(x + 9\right)^{8} \left(9 x + 6\right)^{6}} \right)