Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 6\right)^{6} \left(x + 4\right)^{3} \left(2 x + 6\right)^{4} e^{x}}{\left(2 - 5 x\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \sin^{2}{\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 - 6\right)^{6} \left(x + 4\right)^{3} \left(2 x + 6\right)^{4} e^{x}}{\left(2 - 5 x\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(x - 6 \right)} + 3 \ln{\left(x + 4 \right)} + 4 \ln{\left(2 x + 6 \right)}- 8 \ln{\left(2 - 5 x \right)} - \frac{7 \ln{\left(5 x + 9 \right)}}{2} - 2 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(5 x + 9\right)} + \frac{8}{2 x + 6} + \frac{3}{x + 4} + \frac{6}{x - 6} + \frac{40}{2 - 5 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(5 x + 9\right)} + \frac{8}{2 x + 6} + \frac{3}{x + 4} + \frac{6}{x - 6} + \frac{40}{2 - 5 x}\right)\left(\frac{\left(x - 6\right)^{6} \left(x + 4\right)^{3} \left(2 x + 6\right)^{4} e^{x}}{\left(2 - 5 x\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{8}{2 x + 6} + \frac{3}{x + 4} + \frac{6}{x - 6}- \frac{2}{\tan{\left(x \right)}} - \frac{35}{2 \left(5 x + 9\right)} + \frac{40}{2 - 5 x}\right)\left(\frac{\left(x - 6\right)^{6} \left(x + 4\right)^{3} \left(2 x + 6\right)^{4} e^{x}}{\left(2 - 5 x\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \sin^{2}{\left(x \right)}} \right)