Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 x + 7\right)^{5} \left(9 x - 3\right)^{2} e^{- x}}{\left(5 x + 5\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 + 7\right)^{5} \left(9 x - 3\right)^{2} e^{- x}}{\left(5 x + 5\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) = 5 \ln{\left(6 x + 7 \right)} + 2 \ln{\left(9 x - 3 \right)}- x - 8 \ln{\left(5 x + 5 \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{18}{9 x - 3} + \frac{30}{6 x + 7} - \frac{40}{5 x + 5} 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 - 3} + \frac{30}{6 x + 7} - \frac{40}{5 x + 5}\right)\left(\frac{\left(6 x + 7\right)^{5} \left(9 x - 3\right)^{2} e^{- x}}{\left(5 x + 5\right)^{8} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{18}{9 x - 3} + \frac{30}{6 x + 7}-1 - \frac{5}{\tan{\left(x \right)}} - \frac{40}{5 x + 5}\right)\left(\frac{\left(6 x + 7\right)^{5} \left(9 x - 3\right)^{2} e^{- x}}{\left(5 x + 5\right)^{8} \sin^{5}{\left(x \right)}} \right)