Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 9\right)^{2} \sqrt{\left(3 x + 9\right)^{7}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{8} \left(8 x + 9\right)^{6}}

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 - 9\right)^{2} \sqrt{\left(3 x + 9\right)^{7}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{8} \left(8 x + 9\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x - 9 \right)} + \frac{7 \ln{\left(3 x + 9 \right)}}{2} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 8 \ln{\left(6 x + 1 \right)} - 6 \ln{\left(8 x + 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{48}{8 x + 9} - \frac{48}{6 x + 1} + \frac{21}{2 \left(3 x + 9\right)} + \frac{2}{x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{48}{8 x + 9} - \frac{48}{6 x + 1} + \frac{21}{2 \left(3 x + 9\right)} + \frac{2}{x - 9}\right)\left(\frac{\left(x - 9\right)^{2} \sqrt{\left(3 x + 9\right)^{7}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{8} \left(8 x + 9\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{21}{2 \left(3 x + 9\right)} + \frac{2}{x - 9}-1 - \frac{48}{8 x + 9} - \frac{48}{6 x + 1}\right)\left(\frac{\left(x - 9\right)^{2} \sqrt{\left(3 x + 9\right)^{7}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{8} \left(8 x + 9\right)^{6}} \right)