Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 2 x - 8\right)^{7} \left(x + 2\right)^{4} e^{- x} \cos^{5}{\left(x \right)}}{\left(1 - 5 x\right)^{5} \left(3 x + 2\right)^{5} \sin^{7}{\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(- 2 x - 8\right)^{7} \left(x + 2\right)^{4} e^{- x} \cos^{5}{\left(x \right)}}{\left(1 - 5 x\right)^{5} \left(3 x + 2\right)^{5} \sin^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(- 2 x - 8 \right)} + 4 \ln{\left(x + 2 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 5 \ln{\left(1 - 5 x \right)} - 5 \ln{\left(3 x + 2 \right)} - 7 \ln{\left(\sin{\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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 2} + \frac{4}{x + 2} - \frac{14}{- 2 x - 8} + \frac{25}{1 - 5 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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 2} + \frac{4}{x + 2} - \frac{14}{- 2 x - 8} + \frac{25}{1 - 5 x}\right)\left(\frac{\left(- 2 x - 8\right)^{7} \left(x + 2\right)^{4} e^{- x} \cos^{5}{\left(x \right)}}{\left(1 - 5 x\right)^{5} \left(3 x + 2\right)^{5} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{4}{x + 2} - \frac{14}{- 2 x - 8}-1 - \frac{7}{\tan{\left(x \right)}} - \frac{15}{3 x + 2} + \frac{25}{1 - 5 x}\right)\left(\frac{\left(- 2 x - 8\right)^{7} \left(x + 2\right)^{4} e^{- x} \cos^{5}{\left(x \right)}}{\left(1 - 5 x\right)^{5} \left(3 x + 2\right)^{5} \sin^{7}{\left(x \right)}} \right)