Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 - 3 x\right)^{6} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(x - 6\right)^{5} \left(4 x - 4\right)^{7} \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 - 3 x\right)^{6} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(x - 6\right)^{5} \left(4 x - 4\right)^{7} \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 - 3 x \right)} + \frac{3 \ln{\left(5 x + 2 \right)}}{2}- x - 5 \ln{\left(x - 6 \right)} - 7 \ln{\left(4 x - 4 \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{15}{2 \left(5 x + 2\right)} - \frac{28}{4 x - 4} - \frac{5}{x - 6} - \frac{18}{9 - 3 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{15}{2 \left(5 x + 2\right)} - \frac{28}{4 x - 4} - \frac{5}{x - 6} - \frac{18}{9 - 3 x}\right)\left(\frac{\left(9 - 3 x\right)^{6} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(x - 6\right)^{5} \left(4 x - 4\right)^{7} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{15}{2 \left(5 x + 2\right)} - \frac{18}{9 - 3 x}5 \tan{\left(x \right)} - 1 - \frac{28}{4 x - 4} - \frac{5}{x - 6}\right)\left(\frac{\left(9 - 3 x\right)^{6} \sqrt{\left(5 x + 2\right)^{3}} e^{- x}}{\left(x - 6\right)^{5} \left(4 x - 4\right)^{7} \cos^{5}{\left(x \right)}} \right)