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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(- 7 x - 9\right)^{2} e^{x} \sin^{5}{\left(x \right)}}{\left(8 x + 6\right)^{2} \left(9 x + 7\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(- 7 x - 9 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(8 x + 6 \right)} - 2 \ln{\left(9 x + 7 \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{18}{9 x + 7} - \frac{16}{8 x + 6} - \frac{14}{- 7 x - 9} 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{18}{9 x + 7} - \frac{16}{8 x + 6} - \frac{14}{- 7 x - 9}\right)\left(\frac{\left(- 7 x - 9\right)^{2} e^{x} \sin^{5}{\left(x \right)}}{\left(8 x + 6\right)^{2} \left(9 x + 7\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{5}{\tan{\left(x \right)}} - \frac{14}{- 7 x - 9}- \frac{18}{9 x + 7} - \frac{16}{8 x + 6}\right)\left(\frac{\left(- 7 x - 9\right)^{2} e^{x} \sin^{5}{\left(x \right)}}{\left(8 x + 6\right)^{2} \left(9 x + 7\right)^{2}} \right)