Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 x - 5\right)^{8} e^{- x} \cos^{3}{\left(x \right)}}{\left(x - 9\right)^{4} \left(3 x - 2\right)^{4} \sqrt{\left(3 x + 8\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{\left(8 x - 5\right)^{8} e^{- x} \cos^{3}{\left(x \right)}}{\left(x - 9\right)^{4} \left(3 x - 2\right)^{4} \sqrt{\left(3 x + 8\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(8 x - 5 \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- x - 4 \ln{\left(x - 9 \right)} - 4 \ln{\left(3 x - 2 \right)} - \frac{7 \ln{\left(3 x + 8 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{64}{8 x - 5} - \frac{21}{2 \left(3 x + 8\right)} - \frac{12}{3 x - 2} - \frac{4}{x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{64}{8 x - 5} - \frac{21}{2 \left(3 x + 8\right)} - \frac{12}{3 x - 2} - \frac{4}{x - 9}\right)\left(\frac{\left(8 x - 5\right)^{8} e^{- x} \cos^{3}{\left(x \right)}}{\left(x - 9\right)^{4} \left(3 x - 2\right)^{4} \sqrt{\left(3 x + 8\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{64}{8 x - 5}-1 - \frac{21}{2 \left(3 x + 8\right)} - \frac{12}{3 x - 2} - \frac{4}{x - 9}\right)\left(\frac{\left(8 x - 5\right)^{8} e^{- x} \cos^{3}{\left(x \right)}}{\left(x - 9\right)^{4} \left(3 x - 2\right)^{4} \sqrt{\left(3 x + 8\right)^{7}}} \right)