Find the derivative of $LaTeX: \displaystyle y = \frac{\left(x - 7\right)^{4} \left(3 x + 1\right)^{3} \sin^{7}{\left(x \right)}}{\left(9 x + 2\right)^{3} \sqrt{\left(x + 4\right)^{5}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(x - 7\right)^{4} \left(3 x + 1\right)^{3} \sin^{7}{\left(x \right)}}{\left(9 x + 2\right)^{3} \sqrt{\left(x + 4\right)^{5}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 4 \ln{\left(x - 7 \right)} + 3 \ln{\left(3 x + 1 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- \frac{5 \ln{\left(x + 4 \right)}}{2} - 3 \ln{\left(9 x + 2 \right)}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{27}{9 x + 2} + \frac{9}{3 x + 1} - \frac{5}{2 \left(x + 4\right)} + \frac{4}{x - 7}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(\frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{27}{9 x + 2} + \frac{9}{3 x + 1} - \frac{5}{2 \left(x + 4\right)} + \frac{4}{x - 7}\right)\left(\frac{\left(x - 7\right)^{4} \left(3 x + 1\right)^{3} \sin^{7}{\left(x \right)}}{\left(9 x + 2\right)^{3} \sqrt{\left(x + 4\right)^{5}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{9}{3 x + 1} + \frac{4}{x - 7}- \frac{27}{9 x + 2} - \frac{5}{2 \left(x + 4\right)}\right)\left(\frac{\left(x - 7\right)^{4} \left(3 x + 1\right)^{3} \sin^{7}{\left(x \right)}}{\left(9 x + 2\right)^{3} \sqrt{\left(x + 4\right)^{5}}} \right)$