Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 x - 6\right)^{6} \left(8 x + 3\right)^{8} e^{x}}{\sqrt{6 x + 2} \left(7 x - 6\right)^{5} \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(6 x - 6\right)^{6} \left(8 x + 3\right)^{8} e^{x}}{\sqrt{6 x + 2} \left(7 x - 6\right)^{5} \sin^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(6 x - 6 \right)} + 8 \ln{\left(8 x + 3 \right)}- \frac{\ln{\left(6 x + 2 \right)}}{2} - 5 \ln{\left(7 x - 6 \right)} - 3 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{64}{8 x + 3} - \frac{35}{7 x - 6} - \frac{3}{6 x + 2} + \frac{36}{6 x - 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{64}{8 x + 3} - \frac{35}{7 x - 6} - \frac{3}{6 x + 2} + \frac{36}{6 x - 6}\right)\left(\frac{\left(6 x - 6\right)^{6} \left(8 x + 3\right)^{8} e^{x}}{\sqrt{6 x + 2} \left(7 x - 6\right)^{5} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{64}{8 x + 3} + \frac{36}{6 x - 6}- \frac{3}{\tan{\left(x \right)}} - \frac{35}{7 x - 6} - \frac{3}{6 x + 2}\right)\left(\frac{\left(6 x - 6\right)^{6} \left(8 x + 3\right)^{8} e^{x}}{\sqrt{6 x + 2} \left(7 x - 6\right)^{5} \sin^{3}{\left(x \right)}} \right)