Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 4\right)^{2} \left(x - 1\right)^{4} \left(3 x + 4\right)^{3} e^{x}}{\sqrt{4 x + 9} \left(8 x + 7\right)^{8} \sin^{5}{\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(- 6 x - 4\right)^{2} \left(x - 1\right)^{4} \left(3 x + 4\right)^{3} e^{x}}{\sqrt{4 x + 9} \left(8 x + 7\right)^{8} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(- 6 x - 4 \right)} + 4 \ln{\left(x - 1 \right)} + 3 \ln{\left(3 x + 4 \right)}- \frac{\ln{\left(4 x + 9 \right)}}{2} - 8 \ln{\left(8 x + 7 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{64}{8 x + 7} - \frac{2}{4 x + 9} + \frac{9}{3 x + 4} + \frac{4}{x - 1} - \frac{12}{- 6 x - 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{64}{8 x + 7} - \frac{2}{4 x + 9} + \frac{9}{3 x + 4} + \frac{4}{x - 1} - \frac{12}{- 6 x - 4}\right)\left(\frac{\left(- 6 x - 4\right)^{2} \left(x - 1\right)^{4} \left(3 x + 4\right)^{3} e^{x}}{\sqrt{4 x + 9} \left(8 x + 7\right)^{8} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{9}{3 x + 4} + \frac{4}{x - 1} - \frac{12}{- 6 x - 4}- \frac{5}{\tan{\left(x \right)}} - \frac{64}{8 x + 7} - \frac{2}{4 x + 9}\right)\left(\frac{\left(- 6 x - 4\right)^{2} \left(x - 1\right)^{4} \left(3 x + 4\right)^{3} e^{x}}{\sqrt{4 x + 9} \left(8 x + 7\right)^{8} \sin^{5}{\left(x \right)}} \right)