Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 - 9 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(2 x + 8\right)^{6} \left(3 x + 4\right)^{6} \cos^{4}{\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(9 - 9 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(2 x + 8\right)^{6} \left(3 x + 4\right)^{6} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(9 - 9 x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- x - 6 \ln{\left(2 x + 8 \right)} - 6 \ln{\left(3 x + 4 \right)} - 4 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{3 x + 4} - \frac{12}{2 x + 8} - \frac{36}{9 - 9 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{3 x + 4} - \frac{12}{2 x + 8} - \frac{36}{9 - 9 x}\right)\left(\frac{\left(9 - 9 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(2 x + 8\right)^{6} \left(3 x + 4\right)^{6} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} - \frac{36}{9 - 9 x}4 \tan{\left(x \right)} - 1 - \frac{18}{3 x + 4} - \frac{12}{2 x + 8}\right)\left(\frac{\left(9 - 9 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(2 x + 8\right)^{6} \left(3 x + 4\right)^{6} \cos^{4}{\left(x \right)}} \right)