Find the derivative of LaTeX: \displaystyle y = \frac{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 6\right)^{2} \sqrt{5 x + 8} \left(6 x - 9\right)^{4}}

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^{3}{\left(x \right)}}{\left(x + 6\right)^{2} \sqrt{5 x + 8} \left(6 x - 9\right)^{4}} \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)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x + 6 \right)} - \frac{\ln{\left(5 x + 8 \right)}}{2} - 4 \ln{\left(6 x - 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{24}{6 x - 9} - \frac{5}{2 \left(5 x + 8\right)} - \frac{2}{x + 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{24}{6 x - 9} - \frac{5}{2 \left(5 x + 8\right)} - \frac{2}{x + 6}\right)\left(\frac{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 6\right)^{2} \sqrt{5 x + 8} \left(6 x - 9\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{6}{\tan{\left(x \right)}}- \frac{24}{6 x - 9} - \frac{5}{2 \left(5 x + 8\right)} - \frac{2}{x + 6}\right)\left(\frac{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 6\right)^{2} \sqrt{5 x + 8} \left(6 x - 9\right)^{4}} \right)