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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(- 4 x - 6\right)^{6} \left(8 x + 5\right)^{7} \cos^{6}{\left(x \right)}}{\left(x - 5\right)^{2} \sqrt{\left(8 x + 1\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 6 \ln{\left(- 4 x - 6 \right)} + 7 \ln{\left(8 x + 5 \right)} + 6 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x - 5 \right)} - \frac{3 \ln{\left(8 x + 1 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{56}{8 x + 5} - \frac{12}{8 x + 1} - \frac{2}{x - 5} - \frac{24}{- 4 x - 6}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{56}{8 x + 5} - \frac{12}{8 x + 1} - \frac{2}{x - 5} - \frac{24}{- 4 x - 6}\right)\left(\frac{\left(- 4 x - 6\right)^{6} \left(8 x + 5\right)^{7} \cos^{6}{\left(x \right)}}{\left(x - 5\right)^{2} \sqrt{\left(8 x + 1\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{56}{8 x + 5} - \frac{24}{- 4 x - 6}- \frac{12}{8 x + 1} - \frac{2}{x - 5}\right)\left(\frac{\left(- 4 x - 6\right)^{6} \left(8 x + 5\right)^{7} \cos^{6}{\left(x \right)}}{\left(x - 5\right)^{2} \sqrt{\left(8 x + 1\right)^{3}}} \right)$