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

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 - 6 x\right)^{5} \left(9 x + 4\right)^{4} \sin^{6}{\left(x \right)}}{\left(x - 5\right)^{8} \sqrt{5 x + 9}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(9 - 6 x \right)} + 4 \ln{\left(9 x + 4 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(x - 5 \right)} - \frac{\ln{\left(5 x + 9 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{9 x + 4} - \frac{5}{2 \left(5 x + 9\right)} - \frac{8}{x - 5} - \frac{30}{9 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{9 x + 4} - \frac{5}{2 \left(5 x + 9\right)} - \frac{8}{x - 5} - \frac{30}{9 - 6 x}\right)\left(\frac{\left(9 - 6 x\right)^{5} \left(9 x + 4\right)^{4} \sin^{6}{\left(x \right)}}{\left(x - 5\right)^{8} \sqrt{5 x + 9}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{6}{\tan{\left(x \right)}} + \frac{36}{9 x + 4} - \frac{30}{9 - 6 x}- \frac{5}{2 \left(5 x + 9\right)} - \frac{8}{x - 5}\right)\left(\frac{\left(9 - 6 x\right)^{5} \left(9 x + 4\right)^{4} \sin^{6}{\left(x \right)}}{\left(x - 5\right)^{8} \sqrt{5 x + 9}} \right)