Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 8\right)^{4} e^{- x} \cos^{7}{\left(x \right)}}{\left(3 x - 9\right)^{2} \left(4 x + 1\right)^{5}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x - 8\right)^{4} e^{- x} \cos^{7}{\left(x \right)}}{\left(3 x - 9\right)^{2} \left(4 x + 1\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x - 8 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- x - 2 \ln{\left(3 x - 9 \right)} - 5 \ln{\left(4 x + 1 \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{20}{4 x + 1} - \frac{6}{3 x - 9} + \frac{4}{x - 8} 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{20}{4 x + 1} - \frac{6}{3 x - 9} + \frac{4}{x - 8}\right)\left(\frac{\left(x - 8\right)^{4} e^{- x} \cos^{7}{\left(x \right)}}{\left(3 x - 9\right)^{2} \left(4 x + 1\right)^{5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + \frac{4}{x - 8}-1 - \frac{20}{4 x + 1} - \frac{6}{3 x - 9}\right)\left(\frac{\left(x - 8\right)^{4} e^{- x} \cos^{7}{\left(x \right)}}{\left(3 x - 9\right)^{2} \left(4 x + 1\right)^{5}} \right)