Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - 8 x\right)^{8} \left(- 7 x - 3\right)^{4} e^{x}}{\left(x - 2\right)^{8} \left(9 x + 4\right)^{5} \cos^{5}{\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(4 - 8 x\right)^{8} \left(- 7 x - 3\right)^{4} e^{x}}{\left(x - 2\right)^{8} \left(9 x + 4\right)^{5} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(4 - 8 x \right)} + 4 \ln{\left(- 7 x - 3 \right)}- 8 \ln{\left(x - 2 \right)} - 5 \ln{\left(9 x + 4 \right)} - 5 \ln{\left(\cos{\left(x \right)} \right)} 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{45}{9 x + 4} - \frac{8}{x - 2} - \frac{28}{- 7 x - 3} - \frac{64}{4 - 8 x} 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{45}{9 x + 4} - \frac{8}{x - 2} - \frac{28}{- 7 x - 3} - \frac{64}{4 - 8 x}\right)\left(\frac{\left(4 - 8 x\right)^{8} \left(- 7 x - 3\right)^{4} e^{x}}{\left(x - 2\right)^{8} \left(9 x + 4\right)^{5} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 - \frac{28}{- 7 x - 3} - \frac{64}{4 - 8 x}5 \tan{\left(x \right)} - \frac{45}{9 x + 4} - \frac{8}{x - 2}\right)\left(\frac{\left(4 - 8 x\right)^{8} \left(- 7 x - 3\right)^{4} e^{x}}{\left(x - 2\right)^{8} \left(9 x + 4\right)^{5} \cos^{5}{\left(x \right)}} \right)