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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(9 x - 7\right)^{6} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 9\right)^{8} \left(3 x - 5\right)^{3} \sqrt{\left(4 x + 1\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 6 \ln{\left(9 x - 7 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(x - 9 \right)} - 3 \ln{\left(3 x - 5 \right)} - \frac{3 \ln{\left(4 x + 1 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{54}{9 x - 7} - \frac{6}{4 x + 1} - \frac{9}{3 x - 5} - \frac{8}{x - 9}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{54}{9 x - 7} - \frac{6}{4 x + 1} - \frac{9}{3 x - 5} - \frac{8}{x - 9}\right)\left(\frac{\left(9 x - 7\right)^{6} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 9\right)^{8} \left(3 x - 5\right)^{3} \sqrt{\left(4 x + 1\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} + \frac{54}{9 x - 7}- \frac{6}{4 x + 1} - \frac{9}{3 x - 5} - \frac{8}{x - 9}\right)\left(\frac{\left(9 x - 7\right)^{6} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 9\right)^{8} \left(3 x - 5\right)^{3} \sqrt{\left(4 x + 1\right)^{3}}} \right)$