Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 1\right)^{4} \left(6 x + 1\right)^{4} \sqrt{6 x + 8} e^{- x}}{\left(4 x - 9\right)^{2} \left(5 x + 5\right)^{7} \cos^{7}{\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(x + 1\right)^{4} \left(6 x + 1\right)^{4} \sqrt{6 x + 8} e^{- x}}{\left(4 x - 9\right)^{2} \left(5 x + 5\right)^{7} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x + 1 \right)} + 4 \ln{\left(6 x + 1 \right)} + \frac{\ln{\left(6 x + 8 \right)}}{2}- x - 2 \ln{\left(4 x - 9 \right)} - 7 \ln{\left(5 x + 5 \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{3}{6 x + 8} + \frac{24}{6 x + 1} - \frac{35}{5 x + 5} - \frac{8}{4 x - 9} + \frac{4}{x + 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{3}{6 x + 8} + \frac{24}{6 x + 1} - \frac{35}{5 x + 5} - \frac{8}{4 x - 9} + \frac{4}{x + 1}\right)\left(\frac{\left(x + 1\right)^{4} \left(6 x + 1\right)^{4} \sqrt{6 x + 8} e^{- x}}{\left(4 x - 9\right)^{2} \left(5 x + 5\right)^{7} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{3}{6 x + 8} + \frac{24}{6 x + 1} + \frac{4}{x + 1}7 \tan{\left(x \right)} - 1 - \frac{35}{5 x + 5} - \frac{8}{4 x - 9}\right)\left(\frac{\left(x + 1\right)^{4} \left(6 x + 1\right)^{4} \sqrt{6 x + 8} e^{- x}}{\left(4 x - 9\right)^{2} \left(5 x + 5\right)^{7} \cos^{7}{\left(x \right)}} \right)