Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 8\right)^{5} e^{- x} \cos^{5}{\left(x \right)}}{8 x^{3} \left(9 - 9 x\right)^{2} \left(- 7 x - 3\right)^{2}}

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 - 8\right)^{5} e^{- x} \cos^{5}{\left(x \right)}}{8 x^{3} \left(9 - 9 x\right)^{2} \left(- 7 x - 3\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 8 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 3 \ln{\left(x \right)} - 2 \ln{\left(9 - 9 x \right)} - 2 \ln{\left(- 7 x - 3 \right)} - 3 \ln{\left(2 \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{5}{x - 8} + \frac{14}{- 7 x - 3} + \frac{18}{9 - 9 x} - \frac{3}{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{5}{x - 8} + \frac{14}{- 7 x - 3} + \frac{18}{9 - 9 x} - \frac{3}{x}\right)\left(\frac{\left(x - 8\right)^{5} e^{- x} \cos^{5}{\left(x \right)}}{8 x^{3} \left(9 - 9 x\right)^{2} \left(- 7 x - 3\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{5}{x - 8}-1 + \frac{14}{- 7 x - 3} + \frac{18}{9 - 9 x} - \frac{3}{x}\right)\left(\frac{\left(x - 8\right)^{5} e^{- x} \cos^{5}{\left(x \right)}}{8 x^{3} \left(9 - 9 x\right)^{2} \left(- 7 x - 3\right)^{2}} \right)