Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 1\right)^{8} \left(3 x - 4\right)^{3} e^{x} \cos^{5}{\left(x \right)}}{\left(- 9 x - 5\right)^{2} \sqrt{\left(6 x + 5\right)^{3}}}

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 + 1\right)^{8} \left(3 x - 4\right)^{3} e^{x} \cos^{5}{\left(x \right)}}{\left(- 9 x - 5\right)^{2} \sqrt{\left(6 x + 5\right)^{3}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(x + 1 \right)} + 3 \ln{\left(3 x - 4 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(- 9 x - 5 \right)} - \frac{3 \ln{\left(6 x + 5 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{9}{6 x + 5} + \frac{9}{3 x - 4} + \frac{8}{x + 1} + \frac{18}{- 9 x - 5} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{9}{6 x + 5} + \frac{9}{3 x - 4} + \frac{8}{x + 1} + \frac{18}{- 9 x - 5}\right)\left(\frac{\left(x + 1\right)^{8} \left(3 x - 4\right)^{3} e^{x} \cos^{5}{\left(x \right)}}{\left(- 9 x - 5\right)^{2} \sqrt{\left(6 x + 5\right)^{3}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + 1 + \frac{9}{3 x - 4} + \frac{8}{x + 1}- \frac{9}{6 x + 5} + \frac{18}{- 9 x - 5}\right)\left(\frac{\left(x + 1\right)^{8} \left(3 x - 4\right)^{3} e^{x} \cos^{5}{\left(x \right)}}{\left(- 9 x - 5\right)^{2} \sqrt{\left(6 x + 5\right)^{3}}} \right)