Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 - 8 x\right)^{4} \left(8 x + 2\right)^{6} e^{x}}{\left(8 x + 1\right)^{6} \sqrt{\left(2 x + 8\right)^{3}} \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 - 8 x\right)^{4} \left(8 x + 2\right)^{6} e^{x}}{\left(8 x + 1\right)^{6} \sqrt{\left(2 x + 8\right)^{3}} \sin^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(2 - 8 x \right)} + 6 \ln{\left(8 x + 2 \right)}- \frac{3 \ln{\left(2 x + 8 \right)}}{2} - 6 \ln{\left(8 x + 1 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{48}{8 x + 2} - \frac{48}{8 x + 1} - \frac{3}{2 x + 8} - \frac{32}{2 - 8 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{48}{8 x + 2} - \frac{48}{8 x + 1} - \frac{3}{2 x + 8} - \frac{32}{2 - 8 x}\right)\left(\frac{\left(2 - 8 x\right)^{4} \left(8 x + 2\right)^{6} e^{x}}{\left(8 x + 1\right)^{6} \sqrt{\left(2 x + 8\right)^{3}} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{48}{8 x + 2} - \frac{32}{2 - 8 x}- \frac{7}{\tan{\left(x \right)}} - \frac{48}{8 x + 1} - \frac{3}{2 x + 8}\right)\left(\frac{\left(2 - 8 x\right)^{4} \left(8 x + 2\right)^{6} e^{x}}{\left(8 x + 1\right)^{6} \sqrt{\left(2 x + 8\right)^{3}} \sin^{7}{\left(x \right)}} \right)