Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 - 7 x\right)^{4} \left(x - 3\right)^{4} \left(6 x + 8\right)^{6}}{\sqrt{\left(2 x + 1\right)^{3}} \sin^{3}{\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(7 - 7 x\right)^{4} \left(x - 3\right)^{4} \left(6 x + 8\right)^{6}}{\sqrt{\left(2 x + 1\right)^{3}} \sin^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(7 - 7 x \right)} + 4 \ln{\left(x - 3 \right)} + 6 \ln{\left(6 x + 8 \right)}- \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{6 x + 8} - \frac{3}{2 x + 1} + \frac{4}{x - 3} - \frac{28}{7 - 7 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{36}{6 x + 8} - \frac{3}{2 x + 1} + \frac{4}{x - 3} - \frac{28}{7 - 7 x}\right)\left(\frac{\left(7 - 7 x\right)^{4} \left(x - 3\right)^{4} \left(6 x + 8\right)^{6}}{\sqrt{\left(2 x + 1\right)^{3}} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{36}{6 x + 8} + \frac{4}{x - 3} - \frac{28}{7 - 7 x}- \frac{3}{\tan{\left(x \right)}} - \frac{3}{2 x + 1}\right)\left(\frac{\left(7 - 7 x\right)^{4} \left(x - 3\right)^{4} \left(6 x + 8\right)^{6}}{\sqrt{\left(2 x + 1\right)^{3}} \sin^{3}{\left(x \right)}} \right)