Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 - 3 x\right)^{5} \left(x + 2\right)^{2} \left(2 x - 5\right)^{6} e^{- x}}{\left(- 4 x - 3\right)^{8} \sqrt{\left(6 x + 8\right)^{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(7 - 3 x\right)^{5} \left(x + 2\right)^{2} \left(2 x - 5\right)^{6} e^{- x}}{\left(- 4 x - 3\right)^{8} \sqrt{\left(6 x + 8\right)^{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(7 - 3 x \right)} + 2 \ln{\left(x + 2 \right)} + 6 \ln{\left(2 x - 5 \right)}- x - 8 \ln{\left(- 4 x - 3 \right)} - \frac{7 \ln{\left(6 x + 8 \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{21}{6 x + 8} + \frac{12}{2 x - 5} + \frac{2}{x + 2} + \frac{32}{- 4 x - 3} - \frac{15}{7 - 3 x} 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{21}{6 x + 8} + \frac{12}{2 x - 5} + \frac{2}{x + 2} + \frac{32}{- 4 x - 3} - \frac{15}{7 - 3 x}\right)\left(\frac{\left(7 - 3 x\right)^{5} \left(x + 2\right)^{2} \left(2 x - 5\right)^{6} e^{- x}}{\left(- 4 x - 3\right)^{8} \sqrt{\left(6 x + 8\right)^{7}} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{12}{2 x - 5} + \frac{2}{x + 2} - \frac{15}{7 - 3 x}-1 - \frac{8}{\tan{\left(x \right)}} - \frac{21}{6 x + 8} + \frac{32}{- 4 x - 3}\right)\left(\frac{\left(7 - 3 x\right)^{5} \left(x + 2\right)^{2} \left(2 x - 5\right)^{6} e^{- x}}{\left(- 4 x - 3\right)^{8} \sqrt{\left(6 x + 8\right)^{7}} \sin^{8}{\left(x \right)}} \right)