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