Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 x + 4\right)^{3} \sqrt{\left(9 x + 2\right)^{3}} e^{x}}{\left(x - 6\right)^{2} \sin^{8}{\left(x \right)} \cos^{2}{\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 + 4\right)^{3} \sqrt{\left(9 x + 2\right)^{3}} e^{x}}{\left(x - 6\right)^{2} \sin^{8}{\left(x \right)} \cos^{2}{\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(6 x + 4 \right)} + \frac{3 \ln{\left(9 x + 2 \right)}}{2}- 2 \ln{\left(x - 6 \right)} - 8 \ln{\left(\sin{\left(x \right)} \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{2 \left(9 x + 2\right)} + \frac{18}{6 x + 4} - \frac{2}{x - 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{2 \left(9 x + 2\right)} + \frac{18}{6 x + 4} - \frac{2}{x - 6}\right)\left(\frac{\left(6 x + 4\right)^{3} \sqrt{\left(9 x + 2\right)^{3}} e^{x}}{\left(x - 6\right)^{2} \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{27}{2 \left(9 x + 2\right)} + \frac{18}{6 x + 4}2 \tan{\left(x \right)} - \frac{8}{\tan{\left(x \right)}} - \frac{2}{x - 6}\right)\left(\frac{\left(6 x + 4\right)^{3} \sqrt{\left(9 x + 2\right)^{3}} e^{x}}{\left(x - 6\right)^{2} \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}} \right)