Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 3 x - 9\right)^{7} \left(8 x - 7\right)^{4} \sqrt{\left(7 x + 4\right)^{5}} e^{x}}{\left(x + 3\right)^{6} \left(5 x + 8\right)^{7} \sin^{7}{\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 - 9\right)^{7} \left(8 x - 7\right)^{4} \sqrt{\left(7 x + 4\right)^{5}} e^{x}}{\left(x + 3\right)^{6} \left(5 x + 8\right)^{7} \sin^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(- 3 x - 9 \right)} + \frac{5 \ln{\left(7 x + 4 \right)}}{2} + 4 \ln{\left(8 x - 7 \right)}- 6 \ln{\left(x + 3 \right)} - 7 \ln{\left(5 x + 8 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{32}{8 x - 7} + \frac{35}{2 \left(7 x + 4\right)} - \frac{35}{5 x + 8} - \frac{6}{x + 3} - \frac{21}{- 3 x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{32}{8 x - 7} + \frac{35}{2 \left(7 x + 4\right)} - \frac{35}{5 x + 8} - \frac{6}{x + 3} - \frac{21}{- 3 x - 9}\right)\left(\frac{\left(- 3 x - 9\right)^{7} \left(8 x - 7\right)^{4} \sqrt{\left(7 x + 4\right)^{5}} e^{x}}{\left(x + 3\right)^{6} \left(5 x + 8\right)^{7} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{32}{8 x - 7} + \frac{35}{2 \left(7 x + 4\right)} - \frac{21}{- 3 x - 9}- \frac{7}{\tan{\left(x \right)}} - \frac{35}{5 x + 8} - \frac{6}{x + 3}\right)\left(\frac{\left(- 3 x - 9\right)^{7} \left(8 x - 7\right)^{4} \sqrt{\left(7 x + 4\right)^{5}} e^{x}}{\left(x + 3\right)^{6} \left(5 x + 8\right)^{7} \sin^{7}{\left(x \right)}} \right)