Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 - 9 x\right)^{2} \left(x - 6\right)^{8} \sqrt{\left(2 x + 3\right)^{5}}}{\left(5 x + 9\right)^{6} \cos^{7}{\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 - 9 x\right)^{2} \left(x - 6\right)^{8} \sqrt{\left(2 x + 3\right)^{5}}}{\left(5 x + 9\right)^{6} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(7 - 9 x \right)} + 8 \ln{\left(x - 6 \right)} + \frac{5 \ln{\left(2 x + 3 \right)}}{2}- 6 \ln{\left(5 x + 9 \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{30}{5 x + 9} + \frac{5}{2 x + 3} + \frac{8}{x - 6} - \frac{18}{7 - 9 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{30}{5 x + 9} + \frac{5}{2 x + 3} + \frac{8}{x - 6} - \frac{18}{7 - 9 x}\right)\left(\frac{\left(7 - 9 x\right)^{2} \left(x - 6\right)^{8} \sqrt{\left(2 x + 3\right)^{5}}}{\left(5 x + 9\right)^{6} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{2 x + 3} + \frac{8}{x - 6} - \frac{18}{7 - 9 x}7 \tan{\left(x \right)} - \frac{30}{5 x + 9}\right)\left(\frac{\left(7 - 9 x\right)^{2} \left(x - 6\right)^{8} \sqrt{\left(2 x + 3\right)^{5}}}{\left(5 x + 9\right)^{6} \cos^{7}{\left(x \right)}} \right)