Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 8\right)^{2} \left(8 x + 4\right)^{6}}{\sqrt{9 x + 5} \sin^{7}{\left(x \right)} \cos^{2}{\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(x + 8\right)^{2} \left(8 x + 4\right)^{6}}{\sqrt{9 x + 5} \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x + 8 \right)} + 6 \ln{\left(8 x + 4 \right)}- \frac{\ln{\left(9 x + 5 \right)}}{2} - 7 \ln{\left(\sin{\left(x \right)} \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)} + \frac{48}{8 x + 4} + \frac{2}{x + 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)} + \frac{48}{8 x + 4} + \frac{2}{x + 8}\right)\left(\frac{\left(x + 8\right)^{2} \left(8 x + 4\right)^{6}}{\sqrt{9 x + 5} \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{48}{8 x + 4} + \frac{2}{x + 8}2 \tan{\left(x \right)} - \frac{7}{\tan{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)}\right)\left(\frac{\left(x + 8\right)^{2} \left(8 x + 4\right)^{6}}{\sqrt{9 x + 5} \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}} \right)