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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(1 - 6 x\right)^{4} e^{- x} \cos^{2}{\left(x \right)}}{\left(3 x - 5\right)^{7} \left(4 x - 5\right)^{5} \sqrt{\left(8 x + 8\right)^{7}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 4 \ln{\left(1 - 6 x \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 7 \ln{\left(3 x - 5 \right)} - 5 \ln{\left(4 x - 5 \right)} - \frac{7 \ln{\left(8 x + 8 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{28}{8 x + 8} - \frac{20}{4 x - 5} - \frac{21}{3 x - 5} - \frac{24}{1 - 6 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{28}{8 x + 8} - \frac{20}{4 x - 5} - \frac{21}{3 x - 5} - \frac{24}{1 - 6 x}\right)\left(\frac{\left(1 - 6 x\right)^{4} e^{- x} \cos^{2}{\left(x \right)}}{\left(3 x - 5\right)^{7} \left(4 x - 5\right)^{5} \sqrt{\left(8 x + 8\right)^{7}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 2 \tan{\left(x \right)} - \frac{24}{1 - 6 x}-1 - \frac{28}{8 x + 8} - \frac{20}{4 x - 5} - \frac{21}{3 x - 5}\right)\left(\frac{\left(1 - 6 x\right)^{4} e^{- x} \cos^{2}{\left(x \right)}}{\left(3 x - 5\right)^{7} \left(4 x - 5\right)^{5} \sqrt{\left(8 x + 8\right)^{7}}} \right)$