Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x - 9\right)^{5} \left(8 x - 4\right)^{2} e^{- x} \cos^{7}{\left(x \right)}}{\left(2 - 9 x\right)^{5} \sin^{5}{\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(7 x - 9\right)^{5} \left(8 x - 4\right)^{2} e^{- x} \cos^{7}{\left(x \right)}}{\left(2 - 9 x\right)^{5} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(7 x - 9 \right)} + 2 \ln{\left(8 x - 4 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- x - 5 \ln{\left(2 - 9 x \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{16}{8 x - 4} + \frac{35}{7 x - 9} + \frac{45}{2 - 9 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{16}{8 x - 4} + \frac{35}{7 x - 9} + \frac{45}{2 - 9 x}\right)\left(\frac{\left(7 x - 9\right)^{5} \left(8 x - 4\right)^{2} e^{- x} \cos^{7}{\left(x \right)}}{\left(2 - 9 x\right)^{5} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + \frac{16}{8 x - 4} + \frac{35}{7 x - 9}-1 - \frac{5}{\tan{\left(x \right)}} + \frac{45}{2 - 9 x}\right)\left(\frac{\left(7 x - 9\right)^{5} \left(8 x - 4\right)^{2} e^{- x} \cos^{7}{\left(x \right)}}{\left(2 - 9 x\right)^{5} \sin^{5}{\left(x \right)}} \right)