Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 8\right)^{5} \sqrt{\left(x + 1\right)^{5}} e^{- x}}{\left(1 - 8 x\right)^{4} \sin^{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 + 8\right)^{5} \sqrt{\left(x + 1\right)^{5}} e^{- x}}{\left(1 - 8 x\right)^{4} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{5 \ln{\left(x + 1 \right)}}{2} + 5 \ln{\left(x + 8 \right)}- x - 4 \ln{\left(1 - 8 x \right)} - 8 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{5}{x + 8} + \frac{5}{2 \left(x + 1\right)} + \frac{32}{1 - 8 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{5}{x + 8} + \frac{5}{2 \left(x + 1\right)} + \frac{32}{1 - 8 x}\right)\left(\frac{\left(x + 8\right)^{5} \sqrt{\left(x + 1\right)^{5}} e^{- x}}{\left(1 - 8 x\right)^{4} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{x + 8} + \frac{5}{2 \left(x + 1\right)}-1 - \frac{8}{\tan{\left(x \right)}} + \frac{32}{1 - 8 x}\right)\left(\frac{\left(x + 8\right)^{5} \sqrt{\left(x + 1\right)^{5}} e^{- x}}{\left(1 - 8 x\right)^{4} \sin^{8}{\left(x \right)}} \right)