Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 x - 2\right)^{2} \sqrt{\left(4 x + 6\right)^{7}} \cos^{4}{\left(x \right)}}{\left(x - 6\right)^{7} \left(4 x - 9\right)^{8}}

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 - 2\right)^{2} \sqrt{\left(4 x + 6\right)^{7}} \cos^{4}{\left(x \right)}}{\left(x - 6\right)^{7} \left(4 x - 9\right)^{8}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(4 x + 6 \right)}}{2} + 2 \ln{\left(6 x - 2 \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 7 \ln{\left(x - 6 \right)} - 8 \ln{\left(4 x - 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{12}{6 x - 2} + \frac{14}{4 x + 6} - \frac{32}{4 x - 9} - \frac{7}{x - 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{12}{6 x - 2} + \frac{14}{4 x + 6} - \frac{32}{4 x - 9} - \frac{7}{x - 6}\right)\left(\frac{\left(6 x - 2\right)^{2} \sqrt{\left(4 x + 6\right)^{7}} \cos^{4}{\left(x \right)}}{\left(x - 6\right)^{7} \left(4 x - 9\right)^{8}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{12}{6 x - 2} + \frac{14}{4 x + 6}- \frac{32}{4 x - 9} - \frac{7}{x - 6}\right)\left(\frac{\left(6 x - 2\right)^{2} \sqrt{\left(4 x + 6\right)^{7}} \cos^{4}{\left(x \right)}}{\left(x - 6\right)^{7} \left(4 x - 9\right)^{8}} \right)