Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x + 1\right)^{8} \cos^{4}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \left(x + 3\right)^{6} \sqrt{\left(7 x + 9\right)^{5}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(9 x + 1\right)^{8} \cos^{4}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \left(x + 3\right)^{6} \sqrt{\left(7 x + 9\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(9 x + 1 \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(- 4 x - 9 \right)} - 6 \ln{\left(x + 3 \right)} - \frac{5 \ln{\left(7 x + 9 \right)}}{2} 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{72}{9 x + 1} - \frac{35}{2 \left(7 x + 9\right)} - \frac{6}{x + 3} + \frac{8}{- 4 x - 9} 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{72}{9 x + 1} - \frac{35}{2 \left(7 x + 9\right)} - \frac{6}{x + 3} + \frac{8}{- 4 x - 9}\right)\left(\frac{\left(9 x + 1\right)^{8} \cos^{4}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \left(x + 3\right)^{6} \sqrt{\left(7 x + 9\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{72}{9 x + 1}- \frac{35}{2 \left(7 x + 9\right)} - \frac{6}{x + 3} + \frac{8}{- 4 x - 9}\right)\left(\frac{\left(9 x + 1\right)^{8} \cos^{4}{\left(x \right)}}{\left(- 4 x - 9\right)^{2} \left(x + 3\right)^{6} \sqrt{\left(7 x + 9\right)^{5}}} \right)