Find the derivative of LaTeX: \displaystyle y = \frac{x^{3} \left(3 x + 2\right)^{3} e^{x}}{\sqrt{\left(8 x + 9\right)^{7}} \sin^{3}{\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{x^{3} \left(3 x + 2\right)^{3} e^{x}}{\sqrt{\left(8 x + 9\right)^{7}} \sin^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 3 \ln{\left(x \right)} + 3 \ln{\left(3 x + 2 \right)}- \frac{7 \ln{\left(8 x + 9 \right)}}{2} - 3 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 9} + \frac{9}{3 x + 2} + \frac{3}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 9} + \frac{9}{3 x + 2} + \frac{3}{x}\right)\left(\frac{x^{3} \left(3 x + 2\right)^{3} e^{x}}{\sqrt{\left(8 x + 9\right)^{7}} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{9}{3 x + 2} + \frac{3}{x}- \frac{3}{\tan{\left(x \right)}} - \frac{28}{8 x + 9}\right)\left(\frac{x^{3} \left(3 x + 2\right)^{3} e^{x}}{\sqrt{\left(8 x + 9\right)^{7}} \sin^{3}{\left(x \right)}} \right)