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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(3 - 6 x\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x - 1\right)^{4} \sqrt{\left(8 x + 5\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 7 \ln{\left(3 - 6 x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(x - 1 \right)} - \frac{3 \ln{\left(8 x + 5 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{8 x + 5} - \frac{4}{x - 1} - \frac{42}{3 - 6 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{8 x + 5} - \frac{4}{x - 1} - \frac{42}{3 - 6 x}\right)\left(\frac{\left(3 - 6 x\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x - 1\right)^{4} \sqrt{\left(8 x + 5\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(1 + \frac{2}{\tan{\left(x \right)}} - \frac{42}{3 - 6 x}- \frac{12}{8 x + 5} - \frac{4}{x - 1}\right)\left(\frac{\left(3 - 6 x\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x - 1\right)^{4} \sqrt{\left(8 x + 5\right)^{3}}} \right)$