Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{7} \left(3 x - 6\right)^{7} \sin^{5}{\left(x \right)}}{\left(8 x + 5\right)^{7} \sqrt{\left(6 x + 8\right)^{7}}}

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 + 3\right)^{7} \left(3 x - 6\right)^{7} \sin^{5}{\left(x \right)}}{\left(8 x + 5\right)^{7} \sqrt{\left(6 x + 8\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(x + 3 \right)} + 7 \ln{\left(3 x - 6 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- \frac{7 \ln{\left(6 x + 8 \right)}}{2} - 7 \ln{\left(8 x + 5 \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{56}{8 x + 5} - \frac{21}{6 x + 8} + \frac{21}{3 x - 6} + \frac{7}{x + 3} 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{56}{8 x + 5} - \frac{21}{6 x + 8} + \frac{21}{3 x - 6} + \frac{7}{x + 3}\right)\left(\frac{\left(x + 3\right)^{7} \left(3 x - 6\right)^{7} \sin^{5}{\left(x \right)}}{\left(8 x + 5\right)^{7} \sqrt{\left(6 x + 8\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{21}{3 x - 6} + \frac{7}{x + 3}- \frac{56}{8 x + 5} - \frac{21}{6 x + 8}\right)\left(\frac{\left(x + 3\right)^{7} \left(3 x - 6\right)^{7} \sin^{5}{\left(x \right)}}{\left(8 x + 5\right)^{7} \sqrt{\left(6 x + 8\right)^{7}}} \right)