Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 9\right)^{5} \left(8 x - 3\right)^{2} e^{x}}{\left(2 x - 4\right)^{2} \cos^{6}{\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 + 9\right)^{5} \left(8 x - 3\right)^{2} e^{x}}{\left(2 x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(x + 9 \right)} + 2 \ln{\left(8 x - 3 \right)}- 2 \ln{\left(2 x - 4 \right)} - 6 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{16}{8 x - 3} - \frac{4}{2 x - 4} + \frac{5}{x + 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{16}{8 x - 3} - \frac{4}{2 x - 4} + \frac{5}{x + 9}\right)\left(\frac{\left(x + 9\right)^{5} \left(8 x - 3\right)^{2} e^{x}}{\left(2 x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{16}{8 x - 3} + \frac{5}{x + 9}6 \tan{\left(x \right)} - \frac{4}{2 x - 4}\right)\left(\frac{\left(x + 9\right)^{5} \left(8 x - 3\right)^{2} e^{x}}{\left(2 x - 4\right)^{2} \cos^{6}{\left(x \right)}} \right)