Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x + 8\right)^{5} \sqrt{7 x + 6} \sin^{8}{\left(x \right)}}{\left(8 - 5 x\right)^{5} \left(x + 8\right)^{2}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(5 x + 8\right)^{5} \sqrt{7 x + 6} \sin^{8}{\left(x \right)}}{\left(8 - 5 x\right)^{5} \left(x + 8\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(5 x + 8 \right)} + \frac{\ln{\left(7 x + 6 \right)}}{2} + 8 \ln{\left(\sin{\left(x \right)} \right)}- 5 \ln{\left(8 - 5 x \right)} - 2 \ln{\left(x + 8 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(7 x + 6\right)} + \frac{25}{5 x + 8} - \frac{2}{x + 8} + \frac{25}{8 - 5 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(7 x + 6\right)} + \frac{25}{5 x + 8} - \frac{2}{x + 8} + \frac{25}{8 - 5 x}\right)\left(\frac{\left(5 x + 8\right)^{5} \sqrt{7 x + 6} \sin^{8}{\left(x \right)}}{\left(8 - 5 x\right)^{5} \left(x + 8\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{\tan{\left(x \right)}} + \frac{7}{2 \left(7 x + 6\right)} + \frac{25}{5 x + 8}- \frac{2}{x + 8} + \frac{25}{8 - 5 x}\right)\left(\frac{\left(5 x + 8\right)^{5} \sqrt{7 x + 6} \sin^{8}{\left(x \right)}}{\left(8 - 5 x\right)^{5} \left(x + 8\right)^{2}} \right)