Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{7} \left(x - 8\right)^{5} \left(5 x - 7\right)^{7}}

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)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{7} \left(x - 8\right)^{5} \left(5 x - 7\right)^{7}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x + 8 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 7 \ln{\left(- 4 x - 8 \right)} - 5 \ln{\left(x - 8 \right)} - 7 \ln{\left(5 x - 7 \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 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{5 x - 7} + \frac{2}{x + 8} - \frac{5}{x - 8} + \frac{28}{- 4 x - 8} 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 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{5 x - 7} + \frac{2}{x + 8} - \frac{5}{x - 8} + \frac{28}{- 4 x - 8}\right)\left(\frac{\left(x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{7} \left(x - 8\right)^{5} \left(5 x - 7\right)^{7}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{5}{\tan{\left(x \right)}} + \frac{2}{x + 8}-1 - \frac{35}{5 x - 7} - \frac{5}{x - 8} + \frac{28}{- 4 x - 8}\right)\left(\frac{\left(x + 8\right)^{2} e^{- x} \sin^{5}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 4 x - 8\right)^{7} \left(x - 8\right)^{5} \left(5 x - 7\right)^{7}} \right)