Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 x + 3\right)^{5} \left(5 x + 1\right)^{7} \sqrt{\left(3 x + 5\right)^{5}}}{\left(x - 8\right)^{2} \sin^{5}{\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(2 x + 3\right)^{5} \left(5 x + 1\right)^{7} \sqrt{\left(3 x + 5\right)^{5}}}{\left(x - 8\right)^{2} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(2 x + 3 \right)} + \frac{5 \ln{\left(3 x + 5 \right)}}{2} + 7 \ln{\left(5 x + 1 \right)}- 2 \ln{\left(x - 8 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{5 x + 1} + \frac{15}{2 \left(3 x + 5\right)} + \frac{10}{2 x + 3} - \frac{2}{x - 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{5 x + 1} + \frac{15}{2 \left(3 x + 5\right)} + \frac{10}{2 x + 3} - \frac{2}{x - 8}\right)\left(\frac{\left(2 x + 3\right)^{5} \left(5 x + 1\right)^{7} \sqrt{\left(3 x + 5\right)^{5}}}{\left(x - 8\right)^{2} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{35}{5 x + 1} + \frac{15}{2 \left(3 x + 5\right)} + \frac{10}{2 x + 3}- \frac{5}{\tan{\left(x \right)}} - \frac{2}{x - 8}\right)\left(\frac{\left(2 x + 3\right)^{5} \left(5 x + 1\right)^{7} \sqrt{\left(3 x + 5\right)^{5}}}{\left(x - 8\right)^{2} \sin^{5}{\left(x \right)}} \right)