Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 6\right)^{5} \left(4 x - 1\right)^{6} e^{- x}}{\left(9 - 8 x\right)^{4} \left(3 x - 7\right)^{8} \cos^{4}{\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 - 6\right)^{5} \left(4 x - 1\right)^{6} e^{- x}}{\left(9 - 8 x\right)^{4} \left(3 x - 7\right)^{8} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 6 \right)} + 6 \ln{\left(4 x - 1 \right)}- x - 4 \ln{\left(9 - 8 x \right)} - 8 \ln{\left(3 x - 7 \right)} - 4 \ln{\left(\cos{\left(x \right)} \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{24}{4 x - 1} - \frac{24}{3 x - 7} + \frac{5}{x - 6} + \frac{32}{9 - 8 x} 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{24}{4 x - 1} - \frac{24}{3 x - 7} + \frac{5}{x - 6} + \frac{32}{9 - 8 x}\right)\left(\frac{\left(x - 6\right)^{5} \left(4 x - 1\right)^{6} e^{- x}}{\left(9 - 8 x\right)^{4} \left(3 x - 7\right)^{8} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{24}{4 x - 1} + \frac{5}{x - 6}4 \tan{\left(x \right)} - 1 - \frac{24}{3 x - 7} + \frac{32}{9 - 8 x}\right)\left(\frac{\left(x - 6\right)^{5} \left(4 x - 1\right)^{6} e^{- x}}{\left(9 - 8 x\right)^{4} \left(3 x - 7\right)^{8} \cos^{4}{\left(x \right)}} \right)