Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 - 2 x\right)^{8} \left(x - 2\right)^{3} \left(7 x + 5\right)^{2} e^{- x}}{\left(- 7 x - 8\right)^{5} \sin^{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(2 - 2 x\right)^{8} \left(x - 2\right)^{3} \left(7 x + 5\right)^{2} e^{- x}}{\left(- 7 x - 8\right)^{5} \sin^{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(2 - 2 x \right)} + 3 \ln{\left(x - 2 \right)} + 2 \ln{\left(7 x + 5 \right)}- x - 5 \ln{\left(- 7 x - 8 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{14}{7 x + 5} + \frac{3}{x - 2} + \frac{35}{- 7 x - 8} - \frac{16}{2 - 2 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{14}{7 x + 5} + \frac{3}{x - 2} + \frac{35}{- 7 x - 8} - \frac{16}{2 - 2 x}\right)\left(\frac{\left(2 - 2 x\right)^{8} \left(x - 2\right)^{3} \left(7 x + 5\right)^{2} e^{- x}}{\left(- 7 x - 8\right)^{5} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{14}{7 x + 5} + \frac{3}{x - 2} - \frac{16}{2 - 2 x}-1 - \frac{7}{\tan{\left(x \right)}} + \frac{35}{- 7 x - 8}\right)\left(\frac{\left(2 - 2 x\right)^{8} \left(x - 2\right)^{3} \left(7 x + 5\right)^{2} e^{- x}}{\left(- 7 x - 8\right)^{5} \sin^{7}{\left(x \right)}} \right)