Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 2 x - 3\right)^{4} \left(x - 3\right)^{7} e^{- x}}{\left(6 x + 9\right)^{4} \left(7 x - 6\right)^{3} \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(- 2 x - 3\right)^{4} \left(x - 3\right)^{7} e^{- x}}{\left(6 x + 9\right)^{4} \left(7 x - 6\right)^{3} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(- 2 x - 3 \right)} + 7 \ln{\left(x - 3 \right)}- x - 4 \ln{\left(6 x + 9 \right)} - 3 \ln{\left(7 x - 6 \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{21}{7 x - 6} - \frac{24}{6 x + 9} + \frac{7}{x - 3} - \frac{8}{- 2 x - 3} 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{21}{7 x - 6} - \frac{24}{6 x + 9} + \frac{7}{x - 3} - \frac{8}{- 2 x - 3}\right)\left(\frac{\left(- 2 x - 3\right)^{4} \left(x - 3\right)^{7} e^{- x}}{\left(6 x + 9\right)^{4} \left(7 x - 6\right)^{3} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{x - 3} - \frac{8}{- 2 x - 3}2 \tan{\left(x \right)} - 1 - \frac{21}{7 x - 6} - \frac{24}{6 x + 9}\right)\left(\frac{\left(- 2 x - 3\right)^{4} \left(x - 3\right)^{7} e^{- x}}{\left(6 x + 9\right)^{4} \left(7 x - 6\right)^{3} \cos^{2}{\left(x \right)}} \right)