Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 4 x - 3\right)^{5} \left(5 x + 7\right)^{7} e^{- x}}{\left(3 x - 8\right)^{4} \sqrt{8 x + 7} \sin^{8}{\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(- 4 x - 3\right)^{5} \left(5 x + 7\right)^{7} e^{- x}}{\left(3 x - 8\right)^{4} \sqrt{8 x + 7} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(- 4 x - 3 \right)} + 7 \ln{\left(5 x + 7 \right)}- x - 4 \ln{\left(3 x - 8 \right)} - \frac{\ln{\left(8 x + 7 \right)}}{2} - 8 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 7} + \frac{35}{5 x + 7} - \frac{12}{3 x - 8} - \frac{20}{- 4 x - 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 7} + \frac{35}{5 x + 7} - \frac{12}{3 x - 8} - \frac{20}{- 4 x - 3}\right)\left(\frac{\left(- 4 x - 3\right)^{5} \left(5 x + 7\right)^{7} e^{- x}}{\left(3 x - 8\right)^{4} \sqrt{8 x + 7} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{35}{5 x + 7} - \frac{20}{- 4 x - 3}-1 - \frac{8}{\tan{\left(x \right)}} - \frac{4}{8 x + 7} - \frac{12}{3 x - 8}\right)\left(\frac{\left(- 4 x - 3\right)^{5} \left(5 x + 7\right)^{7} e^{- x}}{\left(3 x - 8\right)^{4} \sqrt{8 x + 7} \sin^{8}{\left(x \right)}} \right)