Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 x + 5\right)^{3} \cos^{7}{\left(x \right)}}{\sqrt{5 x + 4} \left(6 x + 5\right)^{4} \sin^{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(8 x + 5\right)^{3} \cos^{7}{\left(x \right)}}{\sqrt{5 x + 4} \left(6 x + 5\right)^{4} \sin^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(8 x + 5 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- \frac{\ln{\left(5 x + 4 \right)}}{2} - 4 \ln{\left(6 x + 5 \right)} - 2 \ln{\left(\sin{\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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{24}{8 x + 5} - \frac{24}{6 x + 5} - \frac{5}{2 \left(5 x + 4\right)} 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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{24}{8 x + 5} - \frac{24}{6 x + 5} - \frac{5}{2 \left(5 x + 4\right)}\right)\left(\frac{\left(8 x + 5\right)^{3} \cos^{7}{\left(x \right)}}{\sqrt{5 x + 4} \left(6 x + 5\right)^{4} \sin^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + \frac{24}{8 x + 5}- \frac{2}{\tan{\left(x \right)}} - \frac{24}{6 x + 5} - \frac{5}{2 \left(5 x + 4\right)}\right)\left(\frac{\left(8 x + 5\right)^{3} \cos^{7}{\left(x \right)}}{\sqrt{5 x + 4} \left(6 x + 5\right)^{4} \sin^{2}{\left(x \right)}} \right)