Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 2\right)^{4} \left(9 x - 1\right)^{3} e^{x}}{\left(3 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{3}} \cos^{4}{\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 + 2\right)^{4} \left(9 x - 1\right)^{3} e^{x}}{\left(3 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{3}} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(x + 2 \right)} + 3 \ln{\left(9 x - 1 \right)}- 8 \ln{\left(3 x - 1 \right)} - \frac{3 \ln{\left(8 x + 1 \right)}}{2} - 4 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{27}{9 x - 1} - \frac{12}{8 x + 1} - \frac{24}{3 x - 1} + \frac{4}{x + 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{27}{9 x - 1} - \frac{12}{8 x + 1} - \frac{24}{3 x - 1} + \frac{4}{x + 2}\right)\left(\frac{\left(x + 2\right)^{4} \left(9 x - 1\right)^{3} e^{x}}{\left(3 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{3}} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{27}{9 x - 1} + \frac{4}{x + 2}4 \tan{\left(x \right)} - \frac{12}{8 x + 1} - \frac{24}{3 x - 1}\right)\left(\frac{\left(x + 2\right)^{4} \left(9 x - 1\right)^{3} e^{x}}{\left(3 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{3}} \cos^{4}{\left(x \right)}} \right)