Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 - 3 x\right)^{2} \left(x + 3\right)^{5} e^{x}}{\left(- x - 4\right)^{3} \left(4 x + 1\right)^{6} \cos^{2}{\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 - 3 x\right)^{2} \left(x + 3\right)^{5} e^{x}}{\left(- x - 4\right)^{3} \left(4 x + 1\right)^{6} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(3 - 3 x \right)} + 5 \ln{\left(x + 3 \right)}- 3 \ln{\left(- x - 4 \right)} - 6 \ln{\left(4 x + 1 \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{24}{4 x + 1} + \frac{5}{x + 3} + \frac{3}{- x - 4} - \frac{6}{3 - 3 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{24}{4 x + 1} + \frac{5}{x + 3} + \frac{3}{- x - 4} - \frac{6}{3 - 3 x}\right)\left(\frac{\left(3 - 3 x\right)^{2} \left(x + 3\right)^{5} e^{x}}{\left(- x - 4\right)^{3} \left(4 x + 1\right)^{6} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{5}{x + 3} - \frac{6}{3 - 3 x}2 \tan{\left(x \right)} - \frac{24}{4 x + 1} + \frac{3}{- x - 4}\right)\left(\frac{\left(3 - 3 x\right)^{2} \left(x + 3\right)^{5} e^{x}}{\left(- x - 4\right)^{3} \left(4 x + 1\right)^{6} \cos^{2}{\left(x \right)}} \right)