Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 6\right)^{6} \left(7 x - 1\right)^{5} \cos^{8}{\left(x \right)}}{\sqrt{\left(4 x + 6\right)^{5}} \sin^{8}{\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(x - 6\right)^{6} \left(7 x - 1\right)^{5} \cos^{8}{\left(x \right)}}{\sqrt{\left(4 x + 6\right)^{5}} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(x - 6 \right)} + 5 \ln{\left(7 x - 1 \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- \frac{5 \ln{\left(4 x + 6 \right)}}{2} - 8 \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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{7 x - 1} - \frac{10}{4 x + 6} + \frac{6}{x - 6} 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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{7 x - 1} - \frac{10}{4 x + 6} + \frac{6}{x - 6}\right)\left(\frac{\left(x - 6\right)^{6} \left(7 x - 1\right)^{5} \cos^{8}{\left(x \right)}}{\sqrt{\left(4 x + 6\right)^{5}} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{35}{7 x - 1} + \frac{6}{x - 6}- \frac{8}{\tan{\left(x \right)}} - \frac{10}{4 x + 6}\right)\left(\frac{\left(x - 6\right)^{6} \left(7 x - 1\right)^{5} \cos^{8}{\left(x \right)}}{\sqrt{\left(4 x + 6\right)^{5}} \sin^{8}{\left(x \right)}} \right)