Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x - 5\right)^{8} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - x\right)^{8} \left(3 x + 4\right)^{6} \sqrt{6 x + 5}}

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 x - 5\right)^{8} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - x\right)^{8} \left(3 x + 4\right)^{6} \sqrt{6 x + 5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(9 x - 5 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 8 \ln{\left(6 - x \right)} - 6 \ln{\left(3 x + 4 \right)} - \frac{\ln{\left(6 x + 5 \right)}}{2} 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{72}{9 x - 5} - \frac{3}{6 x + 5} - \frac{18}{3 x + 4} + \frac{8}{6 - 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{72}{9 x - 5} - \frac{3}{6 x + 5} - \frac{18}{3 x + 4} + \frac{8}{6 - x}\right)\left(\frac{\left(9 x - 5\right)^{8} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - x\right)^{8} \left(3 x + 4\right)^{6} \sqrt{6 x + 5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{72}{9 x - 5}-1 - \frac{3}{6 x + 5} - \frac{18}{3 x + 4} + \frac{8}{6 - x}\right)\left(\frac{\left(9 x - 5\right)^{8} e^{- x} \cos^{2}{\left(x \right)}}{\left(6 - x\right)^{8} \left(3 x + 4\right)^{6} \sqrt{6 x + 5}} \right)