Find the derivative of $LaTeX: \displaystyle y = \frac{\left(x - 5\right)^{3} \left(x + 1\right)^{7} \cos^{8}{\left(x \right)}}{\left(4 x - 9\right)^{6} \sqrt{\left(4 x + 2\right)^{7}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(x - 5\right)^{3} \left(x + 1\right)^{7} \cos^{8}{\left(x \right)}}{\left(4 x - 9\right)^{6} \sqrt{\left(4 x + 2\right)^{7}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 3 \ln{\left(x - 5 \right)} + 7 \ln{\left(x + 1 \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(4 x - 9 \right)} - \frac{7 \ln{\left(4 x + 2 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{14}{4 x + 2} - \frac{24}{4 x - 9} + \frac{7}{x + 1} + \frac{3}{x - 5}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{14}{4 x + 2} - \frac{24}{4 x - 9} + \frac{7}{x + 1} + \frac{3}{x - 5}\right)\left(\frac{\left(x - 5\right)^{3} \left(x + 1\right)^{7} \cos^{8}{\left(x \right)}}{\left(4 x - 9\right)^{6} \sqrt{\left(4 x + 2\right)^{7}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{7}{x + 1} + \frac{3}{x - 5}- \frac{14}{4 x + 2} - \frac{24}{4 x - 9}\right)\left(\frac{\left(x - 5\right)^{3} \left(x + 1\right)^{7} \cos^{8}{\left(x \right)}}{\left(4 x - 9\right)^{6} \sqrt{\left(4 x + 2\right)^{7}}} \right)$