Find the derivative of LaTeX: \displaystyle y = \frac{6561 x^{8} \sqrt{x + 9} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \left(2 x + 8\right)^{8} \left(8 x - 4\right)^{7}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{6561 x^{8} \sqrt{x + 9} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \left(2 x + 8\right)^{8} \left(8 x - 4\right)^{7}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(x \right)} + \frac{\ln{\left(x + 9 \right)}}{2} + 2 \ln{\left(\cos{\left(x \right)} \right)} + 8 \ln{\left(3 \right)}- x - 8 \ln{\left(x - 3 \right)} - 8 \ln{\left(2 x + 8 \right)} - 7 \ln{\left(8 x - 4 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{56}{8 x - 4} - \frac{16}{2 x + 8} + \frac{1}{2 \left(x + 9\right)} - \frac{8}{x - 3} + \frac{8}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{56}{8 x - 4} - \frac{16}{2 x + 8} + \frac{1}{2 \left(x + 9\right)} - \frac{8}{x - 3} + \frac{8}{x}\right)\left(\frac{6561 x^{8} \sqrt{x + 9} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \left(2 x + 8\right)^{8} \left(8 x - 4\right)^{7}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{1}{2 \left(x + 9\right)} + \frac{8}{x}-1 - \frac{56}{8 x - 4} - \frac{16}{2 x + 8} - \frac{8}{x - 3}\right)\left(\frac{6561 x^{8} \sqrt{x + 9} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \left(2 x + 8\right)^{8} \left(8 x - 4\right)^{7}} \right)