Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 5\right)^{6} \left(9 x + 8\right)^{4} e^{x}}{\left(x + 8\right)^{7} \sin^{8}{\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(x - 5\right)^{6} \left(9 x + 8\right)^{4} e^{x}}{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(x - 5 \right)} + 4 \ln{\left(9 x + 8 \right)}- 7 \ln{\left(x + 8 \right)} - 8 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{9 x + 8} - \frac{7}{x + 8} + \frac{6}{x - 5} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{9 x + 8} - \frac{7}{x + 8} + \frac{6}{x - 5}\right)\left(\frac{\left(x - 5\right)^{6} \left(9 x + 8\right)^{4} e^{x}}{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{36}{9 x + 8} + \frac{6}{x - 5}- \frac{8}{\tan{\left(x \right)}} - \frac{7}{x + 8}\right)\left(\frac{\left(x - 5\right)^{6} \left(9 x + 8\right)^{4} e^{x}}{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}} \right)