Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 2 x - 1\right)^{4} \left(8 x + 4\right)^{2} e^{- x}}{\left(2 x - 3\right)^{7} \sin^{4}{\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(- 2 x - 1\right)^{4} \left(8 x + 4\right)^{2} e^{- x}}{\left(2 x - 3\right)^{7} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(- 2 x - 1 \right)} + 2 \ln{\left(8 x + 4 \right)}- x - 7 \ln{\left(2 x - 3 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{16}{8 x + 4} - \frac{14}{2 x - 3} - \frac{8}{- 2 x - 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{16}{8 x + 4} - \frac{14}{2 x - 3} - \frac{8}{- 2 x - 1}\right)\left(\frac{\left(- 2 x - 1\right)^{4} \left(8 x + 4\right)^{2} e^{- x}}{\left(2 x - 3\right)^{7} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{16}{8 x + 4} - \frac{8}{- 2 x - 1}-1 - \frac{4}{\tan{\left(x \right)}} - \frac{14}{2 x - 3}\right)\left(\frac{\left(- 2 x - 1\right)^{4} \left(8 x + 4\right)^{2} e^{- x}}{\left(2 x - 3\right)^{7} \sin^{4}{\left(x \right)}} \right)