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

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 x - 5\right)^{2} \sqrt{\left(3 x + 6\right)^{5}}}{\left(x + 8\right)^{3} \left(4 x - 4\right)^{3} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{5 \ln{\left(3 x + 6 \right)}}{2} + 2 \ln{\left(9 x - 5 \right)}- 3 \ln{\left(x + 8 \right)} - 3 \ln{\left(4 x - 4 \right)} - 8 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{18}{9 x - 5} - \frac{12}{4 x - 4} + \frac{15}{2 \left(3 x + 6\right)} - \frac{3}{x + 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{18}{9 x - 5} - \frac{12}{4 x - 4} + \frac{15}{2 \left(3 x + 6\right)} - \frac{3}{x + 8}\right)\left(\frac{\left(9 x - 5\right)^{2} \sqrt{\left(3 x + 6\right)^{5}}}{\left(x + 8\right)^{3} \left(4 x - 4\right)^{3} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{18}{9 x - 5} + \frac{15}{2 \left(3 x + 6\right)}- \frac{8}{\tan{\left(x \right)}} - \frac{12}{4 x - 4} - \frac{3}{x + 8}\right)\left(\frac{\left(9 x - 5\right)^{2} \sqrt{\left(3 x + 6\right)^{5}}}{\left(x + 8\right)^{3} \left(4 x - 4\right)^{3} \sin^{8}{\left(x \right)}} \right)