Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 x + 2\right)^{5} \left(5 x + 9\right)^{2}}{\left(x + 8\right)^{3} \sqrt{x + 9} \sin^{4}{\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(3 x + 2\right)^{5} \left(5 x + 9\right)^{2}}{\left(x + 8\right)^{3} \sqrt{x + 9} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(3 x + 2 \right)} + 2 \ln{\left(5 x + 9 \right)}- 3 \ln{\left(x + 8 \right)} - \frac{\ln{\left(x + 9 \right)}}{2} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{10}{5 x + 9} + \frac{15}{3 x + 2} - \frac{1}{2 \left(x + 9\right)} - \frac{3}{x + 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{10}{5 x + 9} + \frac{15}{3 x + 2} - \frac{1}{2 \left(x + 9\right)} - \frac{3}{x + 8}\right)\left(\frac{\left(3 x + 2\right)^{5} \left(5 x + 9\right)^{2}}{\left(x + 8\right)^{3} \sqrt{x + 9} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{10}{5 x + 9} + \frac{15}{3 x + 2}- \frac{4}{\tan{\left(x \right)}} - \frac{1}{2 \left(x + 9\right)} - \frac{3}{x + 8}\right)\left(\frac{\left(3 x + 2\right)^{5} \left(5 x + 9\right)^{2}}{\left(x + 8\right)^{3} \sqrt{x + 9} \sin^{4}{\left(x \right)}} \right)