Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 8 x - 1\right)^{3} \left(x - 7\right)^{6} e^{x}}{\sqrt{\left(4 x + 2\right)^{3}} \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(- 8 x - 1\right)^{3} \left(x - 7\right)^{6} e^{x}}{\sqrt{\left(4 x + 2\right)^{3}} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 3 \ln{\left(- 8 x - 1 \right)} + 6 \ln{\left(x - 7 \right)}- \frac{3 \ln{\left(4 x + 2 \right)}}{2} - 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{6}{4 x + 2} + \frac{6}{x - 7} - \frac{24}{- 8 x - 1} 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{6}{4 x + 2} + \frac{6}{x - 7} - \frac{24}{- 8 x - 1}\right)\left(\frac{\left(- 8 x - 1\right)^{3} \left(x - 7\right)^{6} e^{x}}{\sqrt{\left(4 x + 2\right)^{3}} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{6}{x - 7} - \frac{24}{- 8 x - 1}- \frac{8}{\tan{\left(x \right)}} - \frac{6}{4 x + 2}\right)\left(\frac{\left(- 8 x - 1\right)^{3} \left(x - 7\right)^{6} e^{x}}{\sqrt{\left(4 x + 2\right)^{3}} \sin^{8}{\left(x \right)}} \right)