Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 3 x - 6\right)^{6} \left(8 x + 8\right)^{2} e^{x} \cos^{6}{\left(x \right)}}{\left(x - 2\right)^{5} \sqrt{x + 4}}

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 x - 6\right)^{6} \left(8 x + 8\right)^{2} e^{x} \cos^{6}{\left(x \right)}}{\left(x - 2\right)^{5} \sqrt{x + 4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(- 3 x - 6 \right)} + 2 \ln{\left(8 x + 8 \right)} + 6 \ln{\left(\cos{\left(x \right)} \right)}- 5 \ln{\left(x - 2 \right)} - \frac{\ln{\left(x + 4 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{16}{8 x + 8} - \frac{1}{2 \left(x + 4\right)} - \frac{5}{x - 2} - \frac{18}{- 3 x - 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{16}{8 x + 8} - \frac{1}{2 \left(x + 4\right)} - \frac{5}{x - 2} - \frac{18}{- 3 x - 6}\right)\left(\frac{\left(- 3 x - 6\right)^{6} \left(8 x + 8\right)^{2} e^{x} \cos^{6}{\left(x \right)}}{\left(x - 2\right)^{5} \sqrt{x + 4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + 1 + \frac{16}{8 x + 8} - \frac{18}{- 3 x - 6}- \frac{1}{2 \left(x + 4\right)} - \frac{5}{x - 2}\right)\left(\frac{\left(- 3 x - 6\right)^{6} \left(8 x + 8\right)^{2} e^{x} \cos^{6}{\left(x \right)}}{\left(x - 2\right)^{5} \sqrt{x + 4}} \right)