Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - 6 x\right)^{7} \sqrt{\left(5 x + 2\right)^{7}} e^{- x}}{\left(- 5 x - 6\right)^{2} \left(x - 3\right)^{3} \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(4 - 6 x\right)^{7} \sqrt{\left(5 x + 2\right)^{7}} e^{- x}}{\left(- 5 x - 6\right)^{2} \left(x - 3\right)^{3} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(4 - 6 x \right)} + \frac{7 \ln{\left(5 x + 2 \right)}}{2}- x - 2 \ln{\left(- 5 x - 6 \right)} - 3 \ln{\left(x - 3 \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{35}{2 \left(5 x + 2\right)} - \frac{3}{x - 3} + \frac{10}{- 5 x - 6} - \frac{42}{4 - 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{35}{2 \left(5 x + 2\right)} - \frac{3}{x - 3} + \frac{10}{- 5 x - 6} - \frac{42}{4 - 6 x}\right)\left(\frac{\left(4 - 6 x\right)^{7} \sqrt{\left(5 x + 2\right)^{7}} e^{- x}}{\left(- 5 x - 6\right)^{2} \left(x - 3\right)^{3} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{35}{2 \left(5 x + 2\right)} - \frac{42}{4 - 6 x}-1 - \frac{5}{\tan{\left(x \right)}} - \frac{3}{x - 3} + \frac{10}{- 5 x - 6}\right)\left(\frac{\left(4 - 6 x\right)^{7} \sqrt{\left(5 x + 2\right)^{7}} e^{- x}}{\left(- 5 x - 6\right)^{2} \left(x - 3\right)^{3} \sin^{5}{\left(x \right)}} \right)