Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x - 6\right)^{4} \sqrt{\left(3 x + 9\right)^{7}}}{\left(x + 6\right)^{7} \left(4 x - 5\right)^{4} \cos^{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(5 x - 6\right)^{4} \sqrt{\left(3 x + 9\right)^{7}}}{\left(x + 6\right)^{7} \left(4 x - 5\right)^{4} \cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(3 x + 9 \right)}}{2} + 4 \ln{\left(5 x - 6 \right)}- 7 \ln{\left(x + 6 \right)} - 4 \ln{\left(4 x - 5 \right)} - 8 \ln{\left(\cos{\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{20}{5 x - 6} - \frac{16}{4 x - 5} + \frac{21}{2 \left(3 x + 9\right)} - \frac{7}{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{20}{5 x - 6} - \frac{16}{4 x - 5} + \frac{21}{2 \left(3 x + 9\right)} - \frac{7}{x + 6}\right)\left(\frac{\left(5 x - 6\right)^{4} \sqrt{\left(3 x + 9\right)^{7}}}{\left(x + 6\right)^{7} \left(4 x - 5\right)^{4} \cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{20}{5 x - 6} + \frac{21}{2 \left(3 x + 9\right)}8 \tan{\left(x \right)} - \frac{16}{4 x - 5} - \frac{7}{x + 6}\right)\left(\frac{\left(5 x - 6\right)^{4} \sqrt{\left(3 x + 9\right)^{7}}}{\left(x + 6\right)^{7} \left(4 x - 5\right)^{4} \cos^{8}{\left(x \right)}} \right)