Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 2\right)^{8} \left(5 x + 3\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(2 x - 2\right)^{7} \left(9 x + 2\right)^{2} \cos^{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(x + 2\right)^{8} \left(5 x + 3\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(2 x - 2\right)^{7} \left(9 x + 2\right)^{2} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(x + 2 \right)} + 7 \ln{\left(5 x + 3 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- x - 7 \ln{\left(2 x - 2 \right)} - 2 \ln{\left(9 x + 2 \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{9 x + 2} + \frac{35}{5 x + 3} - \frac{14}{2 x - 2} + \frac{8}{x + 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{9 x + 2} + \frac{35}{5 x + 3} - \frac{14}{2 x - 2} + \frac{8}{x + 2}\right)\left(\frac{\left(x + 2\right)^{8} \left(5 x + 3\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(2 x - 2\right)^{7} \left(9 x + 2\right)^{2} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{\tan{\left(x \right)}} + \frac{35}{5 x + 3} + \frac{8}{x + 2}7 \tan{\left(x \right)} - 1 - \frac{18}{9 x + 2} - \frac{14}{2 x - 2}\right)\left(\frac{\left(x + 2\right)^{8} \left(5 x + 3\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(2 x - 2\right)^{7} \left(9 x + 2\right)^{2} \cos^{7}{\left(x \right)}} \right)