Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 4\right)^{6} \cos^{3}{\left(x \right)}}{49 x^{2} \left(9 - 8 x\right)^{8} \sin^{6}{\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 + 4\right)^{6} \cos^{3}{\left(x \right)}}{49 x^{2} \left(9 - 8 x\right)^{8} \sin^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(x + 4 \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x \right)} - 8 \ln{\left(9 - 8 x \right)} - 6 \ln{\left(\sin{\left(x \right)} \right)} - 2 \ln{\left(7 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{6}{x + 4} + \frac{64}{9 - 8 x} - \frac{2}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{6}{x + 4} + \frac{64}{9 - 8 x} - \frac{2}{x}\right)\left(\frac{\left(x + 4\right)^{6} \cos^{3}{\left(x \right)}}{49 x^{2} \left(9 - 8 x\right)^{8} \sin^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{6}{x + 4}- \frac{6}{\tan{\left(x \right)}} + \frac{64}{9 - 8 x} - \frac{2}{x}\right)\left(\frac{\left(x + 4\right)^{6} \cos^{3}{\left(x \right)}}{49 x^{2} \left(9 - 8 x\right)^{8} \sin^{6}{\left(x \right)}} \right)