Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x + 9\right)^{7} \sqrt{9 x + 2} e^{- x}}{\left(x - 5\right)^{2} \left(x + 5\right)^{6} \cos^{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(5 x + 9\right)^{7} \sqrt{9 x + 2} e^{- x}}{\left(x - 5\right)^{2} \left(x + 5\right)^{6} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(5 x + 9 \right)} + \frac{\ln{\left(9 x + 2 \right)}}{2}- x - 2 \ln{\left(x - 5 \right)} - 6 \ln{\left(x + 5 \right)} - 5 \ln{\left(\cos{\left(x \right)} \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{9}{2 \left(9 x + 2\right)} + \frac{35}{5 x + 9} - \frac{6}{x + 5} - \frac{2}{x - 5} 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{9}{2 \left(9 x + 2\right)} + \frac{35}{5 x + 9} - \frac{6}{x + 5} - \frac{2}{x - 5}\right)\left(\frac{\left(5 x + 9\right)^{7} \sqrt{9 x + 2} e^{- x}}{\left(x - 5\right)^{2} \left(x + 5\right)^{6} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{9}{2 \left(9 x + 2\right)} + \frac{35}{5 x + 9}5 \tan{\left(x \right)} - 1 - \frac{6}{x + 5} - \frac{2}{x - 5}\right)\left(\frac{\left(5 x + 9\right)^{7} \sqrt{9 x + 2} e^{- x}}{\left(x - 5\right)^{2} \left(x + 5\right)^{6} \cos^{5}{\left(x \right)}} \right)