Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x + 4\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(4 - 4 x\right)^{4} \left(6 x - 9\right)^{2} \cos^{2}{\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(5 x + 4\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(4 - 4 x\right)^{4} \left(6 x - 9\right)^{2} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(5 x + 4 \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- x - 4 \ln{\left(4 - 4 x \right)} - 2 \ln{\left(6 x - 9 \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{6 x - 9} + \frac{20}{5 x + 4} + \frac{16}{4 - 4 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{6 x - 9} + \frac{20}{5 x + 4} + \frac{16}{4 - 4 x}\right)\left(\frac{\left(5 x + 4\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(4 - 4 x\right)^{4} \left(6 x - 9\right)^{2} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} + \frac{20}{5 x + 4}2 \tan{\left(x \right)} - 1 - \frac{12}{6 x - 9} + \frac{16}{4 - 4 x}\right)\left(\frac{\left(5 x + 4\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(4 - 4 x\right)^{4} \left(6 x - 9\right)^{2} \cos^{2}{\left(x \right)}} \right)