Find the derivative of LaTeX: \displaystyle y = - \frac{343 x^{3} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \sqrt{9 x + 5} \sin^{4}{\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{343 x^{3} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \sqrt{9 x + 5} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(x \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)} + 3 \ln{\left(7 \right)} + i \pi- 7 \ln{\left(x - 3 \right)} - \frac{\ln{\left(9 x + 5 \right)}}{2} - 4 \ln{\left(\sin{\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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)} - \frac{7}{x - 3} + \frac{3}{x} 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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)} - \frac{7}{x - 3} + \frac{3}{x}\right)\left(- \frac{343 x^{3} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \sqrt{9 x + 5} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{3}{x}- \frac{4}{\tan{\left(x \right)}} - \frac{9}{2 \left(9 x + 5\right)} - \frac{7}{x - 3}\right)\left(- \frac{343 x^{3} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \sqrt{9 x + 5} \sin^{4}{\left(x \right)}} \right)