Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 3 x - 2\right)^{5} \left(3 x - 3\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{25 x^{2} \left(x + 1\right)^{3} \sqrt{x + 2}}

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 - 2\right)^{5} \left(3 x - 3\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{25 x^{2} \left(x + 1\right)^{3} \sqrt{x + 2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(- 3 x - 2 \right)} + 3 \ln{\left(3 x - 3 \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 3 \ln{\left(x + 1 \right)} - \frac{\ln{\left(x + 2 \right)}}{2} - 2 \ln{\left(5 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{9}{3 x - 3} - \frac{1}{2 \left(x + 2\right)} - \frac{3}{x + 1} - \frac{15}{- 3 x - 2} - \frac{2}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{9}{3 x - 3} - \frac{1}{2 \left(x + 2\right)} - \frac{3}{x + 1} - \frac{15}{- 3 x - 2} - \frac{2}{x}\right)\left(\frac{\left(- 3 x - 2\right)^{5} \left(3 x - 3\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{25 x^{2} \left(x + 1\right)^{3} \sqrt{x + 2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + 1 + \frac{9}{3 x - 3} - \frac{15}{- 3 x - 2}- \frac{1}{2 \left(x + 2\right)} - \frac{3}{x + 1} - \frac{2}{x}\right)\left(\frac{\left(- 3 x - 2\right)^{5} \left(3 x - 3\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{25 x^{2} \left(x + 1\right)^{3} \sqrt{x + 2}} \right)