Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 7\right)^{3} \sqrt{\left(8 x + 9\right)^{7}} \sin^{6}{\left(x \right)}}{\left(- 6 x - 2\right)^{3} \left(3 x + 6\right)^{5}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x + 7\right)^{3} \sqrt{\left(8 x + 9\right)^{7}} \sin^{6}{\left(x \right)}}{\left(- 6 x - 2\right)^{3} \left(3 x + 6\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(x + 7 \right)} + \frac{7 \ln{\left(8 x + 9 \right)}}{2} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 3 \ln{\left(- 6 x - 2 \right)} - 5 \ln{\left(3 x + 6 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{8 x + 9} - \frac{15}{3 x + 6} + \frac{3}{x + 7} + \frac{18}{- 6 x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{8 x + 9} - \frac{15}{3 x + 6} + \frac{3}{x + 7} + \frac{18}{- 6 x - 2}\right)\left(\frac{\left(x + 7\right)^{3} \sqrt{\left(8 x + 9\right)^{7}} \sin^{6}{\left(x \right)}}{\left(- 6 x - 2\right)^{3} \left(3 x + 6\right)^{5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{6}{\tan{\left(x \right)}} + \frac{28}{8 x + 9} + \frac{3}{x + 7}- \frac{15}{3 x + 6} + \frac{18}{- 6 x - 2}\right)\left(\frac{\left(x + 7\right)^{3} \sqrt{\left(8 x + 9\right)^{7}} \sin^{6}{\left(x \right)}}{\left(- 6 x - 2\right)^{3} \left(3 x + 6\right)^{5}} \right)