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