Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x + 9\right)^{2} \cos^{7}{\left(x \right)}}{\left(- 4 x - 1\right)^{4} \left(x + 5\right)^{7} \sqrt{\left(6 x + 4\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 + 9\right)^{2} \cos^{7}{\left(x \right)}}{\left(- 4 x - 1\right)^{4} \left(x + 5\right)^{7} \sqrt{\left(6 x + 4\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(9 x + 9 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 4 \ln{\left(- 4 x - 1 \right)} - 7 \ln{\left(x + 5 \right)} - \frac{5 \ln{\left(6 x + 4 \right)}}{2} 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{18}{9 x + 9} - \frac{15}{6 x + 4} - \frac{7}{x + 5} + \frac{16}{- 4 x - 1} 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{18}{9 x + 9} - \frac{15}{6 x + 4} - \frac{7}{x + 5} + \frac{16}{- 4 x - 1}\right)\left(\frac{\left(9 x + 9\right)^{2} \cos^{7}{\left(x \right)}}{\left(- 4 x - 1\right)^{4} \left(x + 5\right)^{7} \sqrt{\left(6 x + 4\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + \frac{18}{9 x + 9}- \frac{15}{6 x + 4} - \frac{7}{x + 5} + \frac{16}{- 4 x - 1}\right)\left(\frac{\left(9 x + 9\right)^{2} \cos^{7}{\left(x \right)}}{\left(- 4 x - 1\right)^{4} \left(x + 5\right)^{7} \sqrt{\left(6 x + 4\right)^{5}}} \right)