Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 - 6 x\right)^{6} \left(4 x - 1\right)^{4} e^{x}}{\left(x + 3\right)^{3} \left(8 x + 2\right)^{5} \sin^{5}{\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(3 - 6 x\right)^{6} \left(4 x - 1\right)^{4} e^{x}}{\left(x + 3\right)^{3} \left(8 x + 2\right)^{5} \sin^{5}{\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(3 - 6 x \right)} + 4 \ln{\left(4 x - 1 \right)}- 3 \ln{\left(x + 3 \right)} - 5 \ln{\left(8 x + 2 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{40}{8 x + 2} + \frac{16}{4 x - 1} - \frac{3}{x + 3} - \frac{36}{3 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{40}{8 x + 2} + \frac{16}{4 x - 1} - \frac{3}{x + 3} - \frac{36}{3 - 6 x}\right)\left(\frac{\left(3 - 6 x\right)^{6} \left(4 x - 1\right)^{4} e^{x}}{\left(x + 3\right)^{3} \left(8 x + 2\right)^{5} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{16}{4 x - 1} - \frac{36}{3 - 6 x}- \frac{5}{\tan{\left(x \right)}} - \frac{40}{8 x + 2} - \frac{3}{x + 3}\right)\left(\frac{\left(3 - 6 x\right)^{6} \left(4 x - 1\right)^{4} e^{x}}{\left(x + 3\right)^{3} \left(8 x + 2\right)^{5} \sin^{5}{\left(x \right)}} \right)