Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 - 6 x\right)^{6} \left(x + 4\right)^{8} \sqrt{\left(8 x + 2\right)^{7}} e^{- x}}{\left(x + 3\right)^{4} \left(5 x + 2\right)^{4} \cos^{5}{\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(9 - 6 x\right)^{6} \left(x + 4\right)^{8} \sqrt{\left(8 x + 2\right)^{7}} e^{- x}}{\left(x + 3\right)^{4} \left(5 x + 2\right)^{4} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(9 - 6 x \right)} + 8 \ln{\left(x + 4 \right)} + \frac{7 \ln{\left(8 x + 2 \right)}}{2}- x - 4 \ln{\left(x + 3 \right)} - 4 \ln{\left(5 x + 2 \right)} - 5 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{28}{8 x + 2} - \frac{20}{5 x + 2} + \frac{8}{x + 4} - \frac{4}{x + 3} - \frac{36}{9 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{28}{8 x + 2} - \frac{20}{5 x + 2} + \frac{8}{x + 4} - \frac{4}{x + 3} - \frac{36}{9 - 6 x}\right)\left(\frac{\left(9 - 6 x\right)^{6} \left(x + 4\right)^{8} \sqrt{\left(8 x + 2\right)^{7}} e^{- x}}{\left(x + 3\right)^{4} \left(5 x + 2\right)^{4} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{28}{8 x + 2} + \frac{8}{x + 4} - \frac{36}{9 - 6 x}5 \tan{\left(x \right)} - 1 - \frac{20}{5 x + 2} - \frac{4}{x + 3}\right)\left(\frac{\left(9 - 6 x\right)^{6} \left(x + 4\right)^{8} \sqrt{\left(8 x + 2\right)^{7}} e^{- x}}{\left(x + 3\right)^{4} \left(5 x + 2\right)^{4} \cos^{5}{\left(x \right)}} \right)