Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 2 x - 4\right)^{3} \left(7 x - 2\right)^{4} \sqrt{\left(6 x + 6\right)^{3}} e^{- x}}{\left(- 3 x - 9\right)^{5} \left(x - 4\right)^{2} \cos^{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(- 2 x - 4\right)^{3} \left(7 x - 2\right)^{4} \sqrt{\left(6 x + 6\right)^{3}} e^{- x}}{\left(- 3 x - 9\right)^{5} \left(x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(- 2 x - 4 \right)} + \frac{3 \ln{\left(6 x + 6 \right)}}{2} + 4 \ln{\left(7 x - 2 \right)}- x - 5 \ln{\left(- 3 x - 9 \right)} - 2 \ln{\left(x - 4 \right)} - 6 \ln{\left(\cos{\left(x \right)} \right)} 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{28}{7 x - 2} + \frac{9}{6 x + 6} - \frac{2}{x - 4} - \frac{6}{- 2 x - 4} + \frac{15}{- 3 x - 9} 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{28}{7 x - 2} + \frac{9}{6 x + 6} - \frac{2}{x - 4} - \frac{6}{- 2 x - 4} + \frac{15}{- 3 x - 9}\right)\left(\frac{\left(- 2 x - 4\right)^{3} \left(7 x - 2\right)^{4} \sqrt{\left(6 x + 6\right)^{3}} e^{- x}}{\left(- 3 x - 9\right)^{5} \left(x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{28}{7 x - 2} + \frac{9}{6 x + 6} - \frac{6}{- 2 x - 4}6 \tan{\left(x \right)} - 1 - \frac{2}{x - 4} + \frac{15}{- 3 x - 9}\right)\left(\frac{\left(- 2 x - 4\right)^{3} \left(7 x - 2\right)^{4} \sqrt{\left(6 x + 6\right)^{3}} e^{- x}}{\left(- 3 x - 9\right)^{5} \left(x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right)