Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 4 x - 4\right)^{7} \left(x - 2\right)^{7} \cos^{8}{\left(x \right)}}{\left(2 x + 1\right)^{6} \sqrt{3 x + 2} \sin^{4}{\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(- 4 x - 4\right)^{7} \left(x - 2\right)^{7} \cos^{8}{\left(x \right)}}{\left(2 x + 1\right)^{6} \sqrt{3 x + 2} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(- 4 x - 4 \right)} + 7 \ln{\left(x - 2 \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(2 x + 1 \right)} - \frac{\ln{\left(3 x + 2 \right)}}{2} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{2 \left(3 x + 2\right)} - \frac{12}{2 x + 1} + \frac{7}{x - 2} - \frac{28}{- 4 x - 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{2 \left(3 x + 2\right)} - \frac{12}{2 x + 1} + \frac{7}{x - 2} - \frac{28}{- 4 x - 4}\right)\left(\frac{\left(- 4 x - 4\right)^{7} \left(x - 2\right)^{7} \cos^{8}{\left(x \right)}}{\left(2 x + 1\right)^{6} \sqrt{3 x + 2} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{7}{x - 2} - \frac{28}{- 4 x - 4}- \frac{4}{\tan{\left(x \right)}} - \frac{3}{2 \left(3 x + 2\right)} - \frac{12}{2 x + 1}\right)\left(\frac{\left(- 4 x - 4\right)^{7} \left(x - 2\right)^{7} \cos^{8}{\left(x \right)}}{\left(2 x + 1\right)^{6} \sqrt{3 x + 2} \sin^{4}{\left(x \right)}} \right)