Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 x + 4\right)^{6} \sqrt{\left(7 x + 9\right)^{5}} e^{- x}}{125 x^{3} \left(- 8 x - 5\right)^{6} \sin^{2}{\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(3 x + 4\right)^{6} \sqrt{\left(7 x + 9\right)^{5}} e^{- x}}{125 x^{3} \left(- 8 x - 5\right)^{6} \sin^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(3 x + 4 \right)} + \frac{5 \ln{\left(7 x + 9 \right)}}{2}- x - 3 \ln{\left(x \right)} - 6 \ln{\left(- 8 x - 5 \right)} - 2 \ln{\left(\sin{\left(x \right)} \right)} - 3 \ln{\left(5 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{2 \left(7 x + 9\right)} + \frac{18}{3 x + 4} + \frac{48}{- 8 x - 5} - \frac{3}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{2 \left(7 x + 9\right)} + \frac{18}{3 x + 4} + \frac{48}{- 8 x - 5} - \frac{3}{x}\right)\left(\frac{\left(3 x + 4\right)^{6} \sqrt{\left(7 x + 9\right)^{5}} e^{- x}}{125 x^{3} \left(- 8 x - 5\right)^{6} \sin^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{35}{2 \left(7 x + 9\right)} + \frac{18}{3 x + 4}-1 - \frac{2}{\tan{\left(x \right)}} + \frac{48}{- 8 x - 5} - \frac{3}{x}\right)\left(\frac{\left(3 x + 4\right)^{6} \sqrt{\left(7 x + 9\right)^{5}} e^{- x}}{125 x^{3} \left(- 8 x - 5\right)^{6} \sin^{2}{\left(x \right)}} \right)