Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 - 2 x\right)^{5} e^{x} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(- 3 x - 8\right)^{3} \sqrt{4 x + 8} \left(8 x + 7\right)^{4}}

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 - 2 x\right)^{5} e^{x} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(- 3 x - 8\right)^{3} \sqrt{4 x + 8} \left(8 x + 7\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(2 - 2 x \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 3 \ln{\left(- 3 x - 8 \right)} - \frac{\ln{\left(4 x + 8 \right)}}{2} - 4 \ln{\left(8 x + 7 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{32}{8 x + 7} - \frac{2}{4 x + 8} + \frac{9}{- 3 x - 8} - \frac{10}{2 - 2 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{32}{8 x + 7} - \frac{2}{4 x + 8} + \frac{9}{- 3 x - 8} - \frac{10}{2 - 2 x}\right)\left(\frac{\left(2 - 2 x\right)^{5} e^{x} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(- 3 x - 8\right)^{3} \sqrt{4 x + 8} \left(8 x + 7\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + 1 + \frac{8}{\tan{\left(x \right)}} - \frac{10}{2 - 2 x}- \frac{32}{8 x + 7} - \frac{2}{4 x + 8} + \frac{9}{- 3 x - 8}\right)\left(\frac{\left(2 - 2 x\right)^{5} e^{x} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(- 3 x - 8\right)^{3} \sqrt{4 x + 8} \left(8 x + 7\right)^{4}} \right)