Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 9\right)^{2} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x + 4\right)^{3} \left(9 x - 8\right)^{2} \sqrt{\left(6 x + 1\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(- 6 x - 9\right)^{2} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x + 4\right)^{3} \left(9 x - 8\right)^{2} \sqrt{\left(6 x + 1\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(- 6 x - 9 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 3 \ln{\left(x + 4 \right)} - \frac{7 \ln{\left(6 x + 1 \right)}}{2} - 2 \ln{\left(9 x - 8 \right)} 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{18}{9 x - 8} - \frac{21}{6 x + 1} - \frac{3}{x + 4} - \frac{12}{- 6 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{18}{9 x - 8} - \frac{21}{6 x + 1} - \frac{3}{x + 4} - \frac{12}{- 6 x - 9}\right)\left(\frac{\left(- 6 x - 9\right)^{2} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x + 4\right)^{3} \left(9 x - 8\right)^{2} \sqrt{\left(6 x + 1\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{6}{\tan{\left(x \right)}} - \frac{12}{- 6 x - 9}- \frac{18}{9 x - 8} - \frac{21}{6 x + 1} - \frac{3}{x + 4}\right)\left(\frac{\left(- 6 x - 9\right)^{2} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(x + 4\right)^{3} \left(9 x - 8\right)^{2} \sqrt{\left(6 x + 1\right)^{7}}} \right)