Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 - 9 x\right)^{4} \left(- 5 x - 2\right)^{6} \sqrt{x + 5} e^{x}}{\left(4 x + 2\right)^{8} \sin^{6}{\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(6 - 9 x\right)^{4} \left(- 5 x - 2\right)^{6} \sqrt{x + 5} e^{x}}{\left(4 x + 2\right)^{8} \sin^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(6 - 9 x \right)} + 6 \ln{\left(- 5 x - 2 \right)} + \frac{\ln{\left(x + 5 \right)}}{2}- 8 \ln{\left(4 x + 2 \right)} - 6 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{32}{4 x + 2} + \frac{1}{2 \left(x + 5\right)} - \frac{30}{- 5 x - 2} - \frac{36}{6 - 9 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{32}{4 x + 2} + \frac{1}{2 \left(x + 5\right)} - \frac{30}{- 5 x - 2} - \frac{36}{6 - 9 x}\right)\left(\frac{\left(6 - 9 x\right)^{4} \left(- 5 x - 2\right)^{6} \sqrt{x + 5} e^{x}}{\left(4 x + 2\right)^{8} \sin^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{1}{2 \left(x + 5\right)} - \frac{30}{- 5 x - 2} - \frac{36}{6 - 9 x}- \frac{6}{\tan{\left(x \right)}} - \frac{32}{4 x + 2}\right)\left(\frac{\left(6 - 9 x\right)^{4} \left(- 5 x - 2\right)^{6} \sqrt{x + 5} e^{x}}{\left(4 x + 2\right)^{8} \sin^{6}{\left(x \right)}} \right)