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

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