Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 3 x - 5\right)^{6} \left(x + 5\right)^{5} e^{- x} \sin^{3}{\left(x \right)}}{\left(3 - 5 x\right)^{6} \left(5 x + 9\right)^{6}}

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 - 5\right)^{6} \left(x + 5\right)^{5} e^{- x} \sin^{3}{\left(x \right)}}{\left(3 - 5 x\right)^{6} \left(5 x + 9\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(- 3 x - 5 \right)} + 5 \ln{\left(x + 5 \right)} + 3 \ln{\left(\sin{\left(x \right)} \right)}- x - 6 \ln{\left(3 - 5 x \right)} - 6 \ln{\left(5 x + 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{30}{5 x + 9} + \frac{5}{x + 5} - \frac{18}{- 3 x - 5} + \frac{30}{3 - 5 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{30}{5 x + 9} + \frac{5}{x + 5} - \frac{18}{- 3 x - 5} + \frac{30}{3 - 5 x}\right)\left(\frac{\left(- 3 x - 5\right)^{6} \left(x + 5\right)^{5} e^{- x} \sin^{3}{\left(x \right)}}{\left(3 - 5 x\right)^{6} \left(5 x + 9\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{3}{\tan{\left(x \right)}} + \frac{5}{x + 5} - \frac{18}{- 3 x - 5}-1 - \frac{30}{5 x + 9} + \frac{30}{3 - 5 x}\right)\left(\frac{\left(- 3 x - 5\right)^{6} \left(x + 5\right)^{5} e^{- x} \sin^{3}{\left(x \right)}}{\left(3 - 5 x\right)^{6} \left(5 x + 9\right)^{6}} \right)