Find the derivative of LaTeX: \displaystyle y = \frac{\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x - 9\right)^{5} \left(5 x - 4\right)^{8} \sqrt{6 x + 5}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x - 9\right)^{5} \left(5 x - 4\right)^{8} \sqrt{6 x + 5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(\sin{\left(x \right)} \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 5 \ln{\left(x - 9 \right)} - 8 \ln{\left(5 x - 4 \right)} - \frac{\ln{\left(6 x + 5 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{6 x + 5} - \frac{40}{5 x - 4} - \frac{5}{x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{6 x + 5} - \frac{40}{5 x - 4} - \frac{5}{x - 9}\right)\left(\frac{\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x - 9\right)^{5} \left(5 x - 4\right)^{8} \sqrt{6 x + 5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{6}{\tan{\left(x \right)}}- \frac{3}{6 x + 5} - \frac{40}{5 x - 4} - \frac{5}{x - 9}\right)\left(\frac{\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x - 9\right)^{5} \left(5 x - 4\right)^{8} \sqrt{6 x + 5}} \right)