Find the derivative of $LaTeX: \displaystyle y = \frac{\left(5 x - 6\right)^{4} \sin^{5}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \sqrt{\left(2 x + 9\right)^{7}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(5 x - 6\right)^{4} \sin^{5}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \sqrt{\left(2 x + 9\right)^{7}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 4 \ln{\left(5 x - 6 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(- 4 x - 9 \right)} - \frac{7 \ln{\left(2 x + 9 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x - 6} - \frac{7}{2 x + 9} + \frac{8}{- 4 x - 9}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(\frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x - 6} - \frac{7}{2 x + 9} + \frac{8}{- 4 x - 9}\right)\left(\frac{\left(5 x - 6\right)^{4} \sin^{5}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \sqrt{\left(2 x + 9\right)^{7}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{20}{5 x - 6}- \frac{7}{2 x + 9} + \frac{8}{- 4 x - 9}\right)\left(\frac{\left(5 x - 6\right)^{4} \sin^{5}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \sqrt{\left(2 x + 9\right)^{7}}} \right)$