Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(8 x + 9\right)^{7}} e^{- x} \cos^{2}{\left(x \right)}}{\left(- x - 8\right)^{3} \left(x + 6\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{\sqrt{\left(8 x + 9\right)^{7}} e^{- x} \cos^{2}{\left(x \right)}}{\left(- x - 8\right)^{3} \left(x + 6\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(8 x + 9 \right)}}{2} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 3 \ln{\left(- x - 8 \right)} - 5 \ln{\left(x + 6 \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{28}{8 x + 9} - \frac{5}{x + 6} + \frac{3}{- x - 8} 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{28}{8 x + 9} - \frac{5}{x + 6} + \frac{3}{- x - 8}\right)\left(\frac{\sqrt{\left(8 x + 9\right)^{7}} e^{- x} \cos^{2}{\left(x \right)}}{\left(- x - 8\right)^{3} \left(x + 6\right)^{5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{28}{8 x + 9}-1 - \frac{5}{x + 6} + \frac{3}{- x - 8}\right)\left(\frac{\sqrt{\left(8 x + 9\right)^{7}} e^{- x} \cos^{2}{\left(x \right)}}{\left(- x - 8\right)^{3} \left(x + 6\right)^{5}} \right)