Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 x + 9\right)^{7} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - 5 x\right)^{4} \left(5 x + 8\right)^{4} \sin^{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(8 x + 9\right)^{7} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - 5 x\right)^{4} \left(5 x + 8\right)^{4} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(8 x + 9 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 4 \ln{\left(6 - 5 x \right)} - 4 \ln{\left(5 x + 8 \right)} - 4 \ln{\left(\sin{\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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{56}{8 x + 9} - \frac{20}{5 x + 8} + \frac{20}{6 - 5 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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{56}{8 x + 9} - \frac{20}{5 x + 8} + \frac{20}{6 - 5 x}\right)\left(\frac{\left(8 x + 9\right)^{7} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - 5 x\right)^{4} \left(5 x + 8\right)^{4} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{56}{8 x + 9}-1 - \frac{4}{\tan{\left(x \right)}} - \frac{20}{5 x + 8} + \frac{20}{6 - 5 x}\right)\left(\frac{\left(8 x + 9\right)^{7} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - 5 x\right)^{4} \left(5 x + 8\right)^{4} \sin^{4}{\left(x \right)}} \right)