Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{5 x + 9} e^{- x} \cos^{4}{\left(x \right)}}{\left(5 x + 7\right)^{5} \left(8 x - 4\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{5 x + 9} e^{- x} \cos^{4}{\left(x \right)}}{\left(5 x + 7\right)^{5} \left(8 x - 4\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{\ln{\left(5 x + 9 \right)}}{2} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 5 \ln{\left(5 x + 7 \right)} - 5 \ln{\left(8 x - 4 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{40}{8 x - 4} + \frac{5}{2 \left(5 x + 9\right)} - \frac{25}{5 x + 7} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{40}{8 x - 4} + \frac{5}{2 \left(5 x + 9\right)} - \frac{25}{5 x + 7}\right)\left(\frac{\sqrt{5 x + 9} e^{- x} \cos^{4}{\left(x \right)}}{\left(5 x + 7\right)^{5} \left(8 x - 4\right)^{5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{5}{2 \left(5 x + 9\right)}-1 - \frac{40}{8 x - 4} - \frac{25}{5 x + 7}\right)\left(\frac{\sqrt{5 x + 9} e^{- x} \cos^{4}{\left(x \right)}}{\left(5 x + 7\right)^{5} \left(8 x - 4\right)^{5}} \right)