Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(x + 2\right)^{5}} e^{- x} \cos^{4}{\left(x \right)}}{\left(3 x + 4\right)^{3} \left(6 x - 4\right)^{8} \left(7 x - 1\right)^{2}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sqrt{\left(x + 2\right)^{5}} e^{- x} \cos^{4}{\left(x \right)}}{\left(3 x + 4\right)^{3} \left(6 x - 4\right)^{8} \left(7 x - 1\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{5 \ln{\left(x + 2 \right)}}{2} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 3 \ln{\left(3 x + 4 \right)} - 8 \ln{\left(6 x - 4 \right)} - 2 \ln{\left(7 x - 1 \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{14}{7 x - 1} - \frac{48}{6 x - 4} - \frac{9}{3 x + 4} + \frac{5}{2 \left(x + 2\right)} 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{14}{7 x - 1} - \frac{48}{6 x - 4} - \frac{9}{3 x + 4} + \frac{5}{2 \left(x + 2\right)}\right)\left(\frac{\sqrt{\left(x + 2\right)^{5}} e^{- x} \cos^{4}{\left(x \right)}}{\left(3 x + 4\right)^{3} \left(6 x - 4\right)^{8} \left(7 x - 1\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{5}{2 \left(x + 2\right)}-1 - \frac{14}{7 x - 1} - \frac{48}{6 x - 4} - \frac{9}{3 x + 4}\right)\left(\frac{\sqrt{\left(x + 2\right)^{5}} e^{- x} \cos^{4}{\left(x \right)}}{\left(3 x + 4\right)^{3} \left(6 x - 4\right)^{8} \left(7 x - 1\right)^{2}} \right)